## A3 - Image and kernel

#### Example 1 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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

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 π

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 $$T$$.

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