A3 - Image and kernel


Example 1 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x + y - 2 \, z \\ x - y \\ -5 \, x + 6 \, y - z \\ -y + z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 1 & -2 \\ 1 & -1 & 0 \\ -5 & 6 & -1 \\ 0 & -1 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 6 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -5 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 6 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 2 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 2 \, x + y - 5 \, z \\ x + y - 2 \, z \\ -3 \, y - 3 \, z \\ -2 \, x + 6 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 2 & 1 & -5 \\ 1 & 1 & -2 \\ 0 & -3 & -3 \\ -2 & 0 & 6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 3 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 0 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 3 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -x - y + 3 \, z \\ -x + 4 \, y - 7 \, z \\ -4 \, y + 8 \, z \\ -x + y - z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & -1 & 3 \\ -1 & 4 & -7 \\ 0 & -4 & 8 \\ -1 & 1 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -4 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 4 \\ -4 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 4 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} -3 \, x_{1} + 3 \, x_{3} \\ -3 \, x_{1} + x_{2} + 4 \, x_{3} \\ -3 \, x_{1} - 3 \, x_{2} \\ 4 \, x_{1} - 4 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -3 & 0 & 3 \\ -3 & 1 & 4 \\ -3 & -3 & 0 \\ 4 & 0 & -4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -3 \\ -3 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ -3 \\ -3 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 5 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x - y + 4 \, z \\ y - 2 \, z \\ -y + 2 \, z \\ -x + 2 \, y - 6 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -1 & 4 \\ 0 & 1 & -2 \\ 0 & -1 & 2 \\ -1 & 2 & -6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -1 \\ 2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -1 \\ 2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 6 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 5 \, x + 4 \, y - 2 \, z \\ -4 \, x - 3 \, y + 2 \, z \\ x + 4 \, y + 6 \, z \\ 4 \, x + 8 \, y + 8 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 5 & 4 & -2 \\ -4 & -3 & 2 \\ 1 & 4 & 6 \\ 4 & 8 & 8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 5 \\ -4 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 4 \\ 8 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 5 \\ -4 \\ 1 \\ 4 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 4 \\ 8 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ -2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 7 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 7 \, x_{2} - 8 \, x_{3} \\ -x_{1} + 4 \, x_{2} + 5 \, x_{3} \\ -2 \, x_{2} - 2 \, x_{3} \\ x_{1} - x_{2} - 2 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -7 & -8 \\ -1 & 4 & 5 \\ 0 & -2 & -2 \\ 1 & -1 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -7 \\ 4 \\ -2 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -7 \\ 4 \\ -2 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 8 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 5 \, x_{2} - 5 \, x_{3} \\ x_{1} + 6 \, x_{2} - 6 \, x_{3} \\ -2 \, x_{2} + 2 \, x_{3} \\ -x_{1} - 7 \, x_{2} + 7 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 5 & -5 \\ 1 & 6 & -6 \\ 0 & -2 & 2 \\ -1 & -7 & 7 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ 6 \\ -2 \\ -7 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ 6 \\ -2 \\ -7 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 9 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -4 \, x + y + 6 \, z \\ -5 \, x + y + 8 \, z \\ -3 \, x + 6 \, z \\ -3 \, x + y + 4 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -4 & 1 & 6 \\ -5 & 1 & 8 \\ -3 & 0 & 6 \\ -3 & 1 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ -5 \\ -3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 10 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 2 \, x - 3 \, y + 4 \, z \\ y - 2 \, z \\ -x + y - z \\ x + y - 3 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 2 & -3 & 4 \\ 0 & 1 & -2 \\ -1 & 1 & -1 \\ 1 & 1 & -3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 1 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 1 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 11 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} -x_{2} + 2 \, x_{3} \\ x_{1} - x_{2} + 5 \, x_{3} \\ 3 \, x_{2} - 6 \, x_{3} \\ -x_{2} + 2 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 0 & -1 & 2 \\ 1 & -1 & 5 \\ 0 & 3 & -6 \\ 0 & -1 & 2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ 3 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ 3 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 12 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 5 \, x_{2} + 2 \, x_{3} \\ 2 \, x_{1} - x_{2} - 7 \, x_{3} \\ x_{1} - 3 \, x_{3} \\ -2 \, x_{2} - 2 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 5 & 2 \\ 2 & -1 & -7 \\ 1 & 0 & -3 \\ 0 & -2 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 5 \\ -1 \\ 0 \\ -2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 3 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 5 \\ -1 \\ 0 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 13 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -4 \, x - 3 \, y - 5 \, z \\ x + 3 \, y - z \\ 5 \, x + 5 \, y + 5 \, z \\ 4 \, x + 5 \, y + 3 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -4 & -3 & -5 \\ 1 & 3 & -1 \\ 5 & 5 & 5 \\ 4 & 5 & 3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 5 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 5 \\ 5 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 5 \\ 4 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 5 \\ 5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 14 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - x_{2} - 4 \, x_{3} \\ x_{2} + 3 \, x_{3} \\ x_{1} - x_{3} \\ -x_{1} + x_{2} + 4 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -1 & -4 \\ 0 & 1 & 3 \\ 1 & 0 & -1 \\ -1 & 1 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -3 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 15 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 4 \, x_{2} - 8 \, x_{3} \\ -x_{1} - 3 \, x_{2} + 6 \, x_{3} \\ 0 \\ 4 \, x_{2} - 8 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 4 & -8 \\ -1 & -3 & 6 \\ 0 & 0 & 0 \\ 0 & 4 & -8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 0 \\ 4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 0 \\ 4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 16 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 2 \, x_{2} + 3 \, x_{3} \\ -2 \, x_{1} + 5 \, x_{2} - 8 \, x_{3} \\ -3 \, x_{1} + 5 \, x_{2} - 7 \, x_{3} \\ x_{2} - 2 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & 3 \\ -2 & 5 & -8 \\ -3 & 5 & -7 \\ 0 & 1 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 5 \\ 5 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -3 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 5 \\ 5 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 17 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 3 \, x - 6 \, y - 6 \, z \\ 2 \, x - 7 \, y - 7 \, z \\ x - y - z \\ -2 \, x + 3 \, y + 3 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 3 & -6 & -6 \\ 2 & -7 & -7 \\ 1 & -1 & -1 \\ -2 & 3 & 3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -6 \\ -7 \\ -1 \\ 3 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ 2 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} -6 \\ -7 \\ -1 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 18 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} -2 \, x_{1} + 5 \, x_{2} + 8 \, x_{3} \\ x_{1} - 3 \, x_{2} - 5 \, x_{3} \\ -x_{2} - 2 \, x_{3} \\ -x_{1} - x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -2 & 5 & 8 \\ 1 & -3 & -5 \\ 0 & -1 & -2 \\ -1 & 0 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ -3 \\ -1 \\ 0 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} 5 \\ -3 \\ -1 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 19 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} 7 \, x_{1} - 4 \, x_{2} + 4 \, x_{3} \\ 2 \, x_{1} - x_{2} + x_{3} \\ 5 \, x_{1} - 6 \, x_{2} + 6 \, x_{3} \\ -5 \, x_{1} + 2 \, x_{2} - 2 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 7 & -4 & 4 \\ 2 & -1 & 1 \\ 5 & -6 & 6 \\ -5 & 2 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 7 \\ 2 \\ 5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -6 \\ 2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 7 \\ 2 \\ 5 \\ -5 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ -6 \\ 2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 20 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} -x_{1} + x_{2} - 2 \, x_{3} \\ 2 \, x_{1} - 3 \, x_{2} + 7 \, x_{3} \\ x_{1} - x_{3} \\ 2 \, x_{1} - 2 \, x_{2} + 4 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & 1 & -2 \\ 2 & -3 & 7 \\ 1 & 0 & -1 \\ 2 & -2 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 21 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 2 \, y - 4 \, z \\ 2 \, x - y + 2 \, z \\ x - 3 \, y + 6 \, z \\ 2 \, x + y - 2 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 0 & 2 & -4 \\ 2 & -1 & 2 \\ 1 & -3 & 6 \\ 2 & 1 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ -3 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ -3 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 22 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 2 \, x - 2 \, y + 6 \, z \\ 2 \, x + y + 3 \, z \\ x - 2 \, y + 4 \, z \\ -x + 5 \, y - 7 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 2 & -2 & 6 \\ 2 & 1 & 3 \\ 1 & -2 & 4 \\ -1 & 5 & -7 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 2 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 5 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ 1 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ -2 \\ 5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 23 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x - 5 \, y + 4 \, z \\ -x - 4 \, y + 5 \, z \\ 5 \, y - 5 \, z \\ 5 \, y - 5 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -5 & 4 \\ -1 & -4 & 5 \\ 0 & 5 & -5 \\ 0 & 5 & -5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ 5 \\ 5 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ -4 \\ 5 \\ 5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 24 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 4 \, x + 4 \, y - 8 \, z \\ 5 \, x + y - 2 \, z \\ 4 \, x - 4 \, y + 8 \, z \\ -3 \, x - 4 \, y + 8 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 4 & 4 & -8 \\ 5 & 1 & -2 \\ 4 & -4 & 8 \\ -3 & -4 & 8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 4 \\ 5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ -4 \\ -4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 5 \\ 4 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ -4 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 25 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -4 \, x + 4 \, y - 4 \, z \\ -2 \, x + y - 4 \, z \\ -5 \, x + 4 \, y - 7 \, z \\ -4 \, x + 3 \, y - 6 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -4 & 4 & -4 \\ -2 & 1 & -4 \\ -5 & 4 & -7 \\ -4 & 3 & -6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -5 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ 4 \\ 3 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ -2 \\ -5 \\ -4 \end{array}\right] , \left[\begin{array}{c} 4 \\ 1 \\ 4 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ -2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 26 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -2 \, x - 5 \, y - z \\ y - z \\ -4 \, y + 4 \, z \\ x + 3 \, y \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -2 & -5 & -1 \\ 0 & 1 & -1 \\ 0 & -4 & 4 \\ 1 & 3 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \\ 3 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ -4 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 27 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 2 \, x_{3} \\ -3 \, x_{1} + x_{2} + 7 \, x_{3} \\ 2 \, x_{1} - 3 \, x_{2} - 7 \, x_{3} \\ 2 \, x_{1} - x_{2} - 5 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 0 & -2 \\ -3 & 1 & 7 \\ 2 & -3 & -7 \\ 2 & -1 & -5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -3 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -3 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ -3 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 28 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x - 7 \, y - 5 \, z \\ x + 3 \, y + 5 \, z \\ -x - 3 \, y - 5 \, z \\ 3 \, y + 3 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -7 & -5 \\ 1 & 3 & 5 \\ -1 & -3 & -5 \\ 0 & 3 & 3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -7 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 29 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} 3 \, x_{1} - 6 \, x_{3} \\ x_{1} - x_{2} - 2 \, x_{3} \\ x_{1} + x_{2} - 2 \, x_{3} \\ -2 \, x_{1} + 7 \, x_{2} + 4 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 3 & 0 & -6 \\ 1 & -1 & -2 \\ 1 & 1 & -2 \\ -2 & 7 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 7 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 7 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 30 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 5 \, x_{2} + 3 \, x_{3} \\ x_{2} - x_{3} \\ -x_{1} + 8 \, x_{2} - 6 \, x_{3} \\ x_{1} - 8 \, x_{2} + 6 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -5 & 3 \\ 0 & 1 & -1 \\ -1 & 8 & -6 \\ 1 & -8 & 6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 8 \\ -8 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 8 \\ -8 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 31 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 4 \, x_{2} - 3 \, x_{3} \\ x_{3} \\ -x_{1} + 4 \, x_{2} + 4 \, x_{3} \\ x_{1} - 4 \, x_{2} - x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -4 & -3 \\ 0 & 0 & 1 \\ -1 & 4 & 4 \\ 1 & -4 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & -4 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 4 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ 4 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 0 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 32 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 4 \, x - 4 \, y + 4 \, z \\ x - 3 \, y + 5 \, z \\ x - z \\ x - 2 \, y + 3 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 4 & -4 & 4 \\ 1 & -3 & 5 \\ 1 & 0 & -1 \\ 1 & -2 & 3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -3 \\ 0 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 33 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 8 \, x_{2} + 2 \, x_{3} \\ -x_{1} + 5 \, x_{2} - 2 \, x_{3} \\ -4 \, x_{2} \\ -5 \, x_{2} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 8 & 2 \\ -1 & 5 & -2 \\ 0 & -4 & 0 \\ 0 & -5 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 8 \\ 5 \\ -4 \\ -5 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 8 \\ 5 \\ -4 \\ -5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 34 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} 5 \, x - 8 \, y - z \\ x - y + z \\ -4 \, x + 6 \, y \\ 3 \, x - 4 \, y + z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 5 & -8 & -1 \\ 1 & -1 & 1 \\ -4 & 6 & 0 \\ 3 & -4 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 5 \\ 1 \\ -4 \\ 3 \end{array}\right] , \left[\begin{array}{c} -8 \\ -1 \\ 6 \\ -4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 5 \\ 1 \\ -4 \\ 3 \end{array}\right] , \left[\begin{array}{c} -8 \\ -1 \\ 6 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ -2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 35 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 4 \, x_{2} + 2 \, x_{3} \\ x_{1} - 4 \, x_{2} + 3 \, x_{3} \\ 2 \, x_{1} - 8 \, x_{2} + 7 \, x_{3} \\ 4 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -4 & 2 \\ 1 & -4 & 3 \\ 2 & -8 & 7 \\ 0 & 0 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & -4 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 3 \\ 7 \\ 4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 3 \\ 7 \\ 4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 0 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 36 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x + 2 \, z \\ 2 \, x + y + z \\ 3 \, x + 2 \, y \\ 0 \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 0 & 2 \\ 2 & 1 & 1 \\ 3 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 2 \\ 0 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 2 \\ 0 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 3 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 37 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -x - 3 \, z \\ y - 3 \, z \\ -x - 3 \, z \\ 2 \, x + y + 3 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & 0 & -3 \\ 0 & 1 & -3 \\ -1 & 0 & -3 \\ 2 & 1 & 3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 3 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ 0 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 0 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 3 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 38 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} 2 \, x_{1} + x_{2} - 3 \, x_{3} \\ x_{1} + x_{2} - x_{3} \\ 2 \, x_{1} + x_{2} - 3 \, x_{3} \\ 2 \, x_{1} - x_{2} - 5 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 2 & 1 & -3 \\ 1 & 1 & -1 \\ 2 & 1 & -3 \\ 2 & -1 & -5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ 1 \\ 2 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 1 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 39 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -2 \, x + y + z \\ y + z \\ -x + 2 \, y + 2 \, z \\ -2 \, y - 2 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -2 & 1 & 1 \\ 0 & 1 & 1 \\ -1 & 2 & 2 \\ 0 & -2 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -2 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ -2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 0 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 2 \\ -2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 40 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - x_{2} \\ 3 \, x_{1} - 2 \, x_{2} + x_{3} \\ -2 \, x_{1} + 6 \, x_{2} + 4 \, x_{3} \\ x_{2} + x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -1 & 0 \\ 3 & -2 & 1 \\ -2 & 6 & 4 \\ 0 & 1 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 6 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -2 \\ 6 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 41 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + x_{2} - 4 \, x_{3} \\ -2 \, x_{1} - x_{2} + 5 \, x_{3} \\ -x_{1} + x_{2} - 2 \, x_{3} \\ -x_{1} + x_{2} - 2 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 1 & -4 \\ -2 & -1 & 5 \\ -1 & 1 & -2 \\ -1 & 1 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 1 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 1 \\ -1 \\ 1 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 42 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} -x_{1} - 5 \, x_{2} - 2 \, x_{3} \\ x_{1} + 4 \, x_{2} + x_{3} \\ 0 \\ x_{1} + 6 \, x_{2} + 3 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & -5 & -2 \\ 1 & 4 & 1 \\ 0 & 0 & 0 \\ 1 & 6 & 3 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ 0 \\ 6 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 3 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 4 \\ 0 \\ 6 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 3 \\ -1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 43 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -y + z \\ y - z \\ x - 6 \, y + 8 \, z \\ x - 3 \, y + 5 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 0 & -1 & 1 \\ 0 & 1 & -1 \\ 1 & -6 & 8 \\ 1 & -3 & 5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -6 \\ -3 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -6 \\ -3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 44 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} - 4 \, x_{2} - 5 \, x_{3} \\ x_{3} \\ -x_{1} + 4 \, x_{2} + 5 \, x_{3} \\ x_{1} - 4 \, x_{2} - 8 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -4 & -5 \\ 0 & 0 & 1 \\ -1 & 4 & 5 \\ 1 & -4 & -8 \end{array}\right] = \left[\begin{array}{ccc} 1 & -4 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 5 \\ -8 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] , \left[\begin{array}{c} -5 \\ 1 \\ 5 \\ -8 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 4 \\ 1 \\ 0 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 45 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 3 \, x_{2} - 5 \, x_{3} \\ -x_{1} - 3 \, x_{2} + 6 \, x_{3} \\ -2 \, x_{1} - 6 \, x_{2} + 5 \, x_{3} \\ x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 3 & -5 \\ -1 & -3 & 6 \\ -2 & -6 & 5 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ 6 \\ 5 \\ 1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -3 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ -2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -5 \\ 6 \\ 5 \\ 1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -3 \\ 1 \\ 0 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 46 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 3 \, x_{2} - 5 \, x_{3} \\ x_{2} - 2 \, x_{3} \\ -x_{1} - x_{2} + x_{3} \\ -x_{1} - x_{2} + x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 3 & -5 \\ 0 & 1 & -2 \\ -1 & -1 & 1 \\ -1 & -1 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -1 \\ -1 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ -1 \end{array}\right] , \left[\begin{array}{c} 3 \\ 1 \\ -1 \\ -1 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 47 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -4 \, y + 8 \, z \\ x - y + 4 \, z \\ -2 \, x + 2 \, y - 8 \, z \\ x + 2 \, y - 2 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 0 & -4 & 8 \\ 1 & -1 & 4 \\ -2 & 2 & -8 \\ 1 & 2 & -2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 0 \\ 1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 2 \\ 2 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 2 \\ 2 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 48 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} x_{1} + 3 \, x_{2} - 4 \, x_{3} \\ x_{1} + 4 \, x_{2} - 6 \, x_{3} \\ 3 \, x_{1} + 5 \, x_{2} - 4 \, x_{3} \\ -2 \, x_{1} - 4 \, x_{2} + 4 \, x_{3} \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & 3 & -4 \\ 1 & 4 & -6 \\ 3 & 5 & -4 \\ -2 & -4 & 4 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ 5 \\ -4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 3 \\ -2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 4 \\ 5 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -2 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 49 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -x + z \\ -x - y + 3 \, z \\ y - 2 \, z \\ 4 \, y - 8 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & 0 & 1 \\ -1 & -1 & 3 \\ 0 & 1 & -2 \\ 0 & 4 & -8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.

Example 50 πŸ”—

Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x - 2 \, y - 5 \, z \\ -x + 3 \, y + 7 \, z \\ 2 \, x - 3 \, y - 8 \, z \\ 3 \, y + 6 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} 1 & -2 & -5 \\ -1 & 3 & 7 \\ 2 & -3 & -8 \\ 0 & 3 & 6 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 3 \\ -3 \\ 3 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.