V9 - Homogeneous systems


Example 1 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} & & & & x_{3} & & &+& x_{5} &=& 0 \\-x_{1} &-& 5 \, x_{2} &+& 5 \, x_{3} &+& 2 \, x_{4} &+& 7 \, x_{5} &=& 0 \\x_{1} &+& 5 \, x_{2} &-& x_{3} &-& 2 \, x_{4} &-& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} -5 \, a + 2 \, b + 2 \, c \\ a \\ -c \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -5 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ -1 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 2 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} 3 \, x_{1} &+& 7 \, x_{2} &-& 9 \, x_{3} &+& 7 \, x_{4} &=& 0 \\4 \, x_{1} &+& 9 \, x_{2} &-& 10 \, x_{3} &+& 6 \, x_{4} &=& 0 \\-2 \, x_{1} &-& 5 \, x_{2} &+& 7 \, x_{3} &-& 6 \, x_{4} &=& 0 \\-x_{1} & & &+& x_{3} &-& 3 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 3 & 7 & -9 & 7 & 0 \\ 4 & 9 & -10 & 6 & 0 \\ -2 & -5 & 7 & -6 & 0 \\ -1 & 0 & 1 & -3 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -2 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ 2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 3 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &+& x_{2} &-& 2 \, x_{3} &+& 7 \, x_{4} &=& 0 \\x_{1} &-& x_{2} &-& 5 \, x_{3} &+& 9 \, x_{4} &=& 0 \\ &-& 3 \, x_{2} &-& 5 \, x_{3} &+& 4 \, x_{4} &=& 0 \\ & & 2 \, x_{2} &-& x_{3} &+& 6 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & 1 & -2 & 7 & 0 \\ 1 & -1 & -5 & 9 & 0 \\ 0 & -3 & -5 & 4 & 0 \\ 0 & 2 & -1 & 6 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ -2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 4 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} 4 \, x_{1} &-& 6 \, x_{2} &-& 6 \, x_{3} &=& 0 \\4 \, x_{1} &-& 11 \, x_{2} &-& x_{3} &=& 0 \\ & & 4 \, x_{2} &-& 4 \, x_{3} &=& 0 \\x_{1} &-& x_{2} &-& 2 \, x_{3} &=& 0 \\-5 \, x_{1} &+& 12 \, x_{2} &+& 3 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 4 & -6 & -6 & 0 \\ 4 & -11 & -1 & 0 \\ 0 & 4 & -4 & 0 \\ 1 & -1 & -2 & 0 \\ -5 & 12 & 3 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -3 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 3 \, a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 5 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &-& 5 \, x_{2} &+& 6 \, x_{3} &=& 0 \\x_{1} & & &+& x_{3} &=& 0 \\4 \, x_{1} &+& x_{2} &+& 3 \, x_{3} &=& 0 \\-5 \, x_{1} &+& 4 \, x_{2} &-& 9 \, x_{3} &=& 0 \\ & & 2 \, x_{2} &-& 2 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 1 & -5 & 6 & 0 \\ 1 & 0 & 1 & 0 \\ 4 & 1 & 3 & 0 \\ -5 & 4 & -9 & 0 \\ 0 & 2 & -2 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 6 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} & & x_{2} & & &-& 2 \, x_{4} &+& 2 \, x_{5} &=& 0 \\-x_{1} &-& x_{2} &+& 3 \, x_{3} &+& 9 \, x_{4} &+& 7 \, x_{5} &=& 0 \\ & & & & x_{3} &+& 2 \, x_{4} &+& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 0 & 1 & 0 & -2 & 2 & 0 \\ -1 & -1 & 3 & 9 & 7 & 0 \\ 0 & 0 & 1 & 2 & 3 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & -2 & 2 & 0 \\ 0 & 0 & 1 & 2 & 3 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} a \\ 2 \, a - 2 \, b \\ -2 \, a - 3 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 1 \\ 2 \\ -2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ -2 \\ -3 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 7 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &-& 2 \, x_{2} &-& 7 \, x_{3} &+& x_{4} &=& 0 \\-x_{1} &-& x_{2} &+& x_{3} &+& 5 \, x_{4} &=& 0 \\ & & 2 \, x_{2} &+& 4 \, x_{3} &-& 4 \, x_{4} &=& 0 \\2 \, x_{1} &+& 2 \, x_{2} &-& 2 \, x_{3} &-& 10 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & -2 & -7 & 1 & 0 \\ -1 & -1 & 1 & 5 & 0 \\ 0 & 2 & 4 & -4 & 0 \\ 2 & 2 & -2 & -10 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & -3 & -3 & 0 \\ 0 & 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 3 \, a + 3 \, b \\ -2 \, a + 2 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 3 \\ -2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 3 \\ 2 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 8 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &-& x_{2} &+& 6 \, x_{3} &+& 6 \, x_{4} &=& 0 \\-x_{1} &-& 2 \, x_{2} &+& 7 \, x_{3} &+& 4 \, x_{4} &=& 0 \\ & & 2 \, x_{2} &-& 9 \, x_{3} &-& 7 \, x_{4} &=& 0 \\-x_{1} &-& x_{2} &+& 3 \, x_{3} &+& x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & -1 & 6 & 6 & 0 \\ -1 & -2 & 7 & 4 & 0 \\ 0 & 2 & -9 & -7 & 0 \\ -1 & -1 & 3 & 1 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ -a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ -1 \\ 1 \end{array}\right] \right\} \).

Example 9 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} 3 \, x_{1} &-& 2 \, x_{2} &-& 3 \, x_{3} &+& 3 \, x_{4} &+& 8 \, x_{5} &=& 0 \\-2 \, x_{1} &-& 3 \, x_{2} &+& 2 \, x_{3} &+& 11 \, x_{4} &-& x_{5} &=& 0 \\-x_{1} &+& 2 \, x_{2} &+& x_{3} &-& 5 \, x_{4} &-& 4 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 3 & -2 & -3 & 3 & 8 & 0 \\ -2 & -3 & 2 & 11 & -1 & 0 \\ -1 & 2 & 1 & -5 & -4 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & -1 & -1 & 2 & 0 \\ 0 & 1 & 0 & -3 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} a + b - 2 \, c \\ 3 \, b + c \\ a \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 3 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 10 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} 2 \, x_{1} &+& 3 \, x_{2} &-& 6 \, x_{3} &=& 0 \\-x_{1} &-& 3 \, x_{2} &+& 6 \, x_{3} &=& 0 \\-2 \, x_{1} &-& 6 \, x_{2} &+& 12 \, x_{3} &=& 0 \\-x_{1} &+& x_{2} &-& 2 \, x_{3} &=& 0 \\-x_{1} &-& x_{2} &+& 2 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 2 & 3 & -6 & 0 \\ -1 & -3 & 6 & 0 \\ -2 & -6 & 12 & 0 \\ -1 & 1 & -2 & 0 \\ -1 & -1 & 2 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 11 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &+& 4 \, x_{2} &-& 5 \, x_{3} &+& 5 \, x_{4} &-& 6 \, x_{5} &=& 0 \\ & & & & & & x_{4} &-& x_{5} &=& 0 \\x_{1} &+& 4 \, x_{2} &-& 5 \, x_{3} &-& x_{4} & & &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & 4 & -5 & 5 & -6 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 \\ 1 & 4 & -5 & -1 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 4 & -5 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -4 \, a + 5 \, b + c \\ a \\ b \\ c \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 5 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 12 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} 7 \, x_{1} &+& 5 \, x_{2} &-& 5 \, x_{3} &-& 7 \, x_{4} &-& 3 \, x_{5} &=& 0 \\-3 \, x_{1} &-& 2 \, x_{2} &+& 2 \, x_{3} &+& 3 \, x_{4} &+& x_{5} &=& 0 \\-3 \, x_{1} &-& 5 \, x_{2} &+& 6 \, x_{3} & & &+& 10 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 7 & 5 & -5 & -7 & -3 & 0 \\ -3 & -2 & 2 & 3 & 1 & 0 \\ -3 & -5 & 6 & 0 & 10 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & -1 & 1 & 0 \\ 0 & 1 & 0 & -3 & 1 & 0 \\ 0 & 0 & 1 & -3 & 3 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} a - b \\ 3 \, a - b \\ 3 \, a - 3 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 1 \\ 3 \\ 3 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -1 \\ -3 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 13 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} & & &-& 3 \, x_{3} &-& 8 \, x_{4} &+& 4 \, x_{5} &=& 0 \\-x_{1} &+& x_{2} &+& 3 \, x_{3} &+& 6 \, x_{4} &-& 3 \, x_{5} &=& 0 \\ &-& x_{2} &+& x_{3} &+& 4 \, x_{4} &-& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} 2 \, a + 2 \, b \\ 2 \, a - b \\ -2 \, a + 2 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 2 \\ 2 \\ -2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 2 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 14 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} -2 \, x_{1} &-& 5 \, x_{2} & & &=& 0 \\ & & x_{2} & & &=& 0 \\ & & 3 \, x_{2} & & &=& 0 \\-x_{1} &+& 8 \, x_{2} & & &=& 0 \\x_{1} & & & & &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 15 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &+& 3 \, x_{2} &+& 4 \, x_{3} &=& 0 \\-2 \, x_{1} &-& 7 \, x_{2} &-& 9 \, x_{3} &=& 0 \\2 \, x_{1} &+& 7 \, x_{2} &+& 9 \, x_{3} &=& 0 \\ &-& x_{2} &-& x_{3} &=& 0 \\ & & x_{2} &+& x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 1 & 3 & 4 & 0 \\ -2 & -7 & -9 & 0 \\ 2 & 7 & 9 & 0 \\ 0 & -1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 1 \end{array}\right] \right\} \).

Example 16 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} -x_{1} &-& 5 \, x_{2} &-& 3 \, x_{3} &-& 10 \, x_{4} &+& 8 \, x_{5} &=& 0 \\x_{1} &+& 5 \, x_{2} &+& 2 \, x_{3} &+& 7 \, x_{4} &-& 6 \, x_{5} &=& 0 \\2 \, x_{1} &+& 10 \, x_{2} &-& 2 \, x_{3} &-& 4 \, x_{4} & & &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} -1 & -5 & -3 & -10 & 8 & 0 \\ 1 & 5 & 2 & 7 & -6 & 0 \\ 2 & 10 & -2 & -4 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 5 & 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 3 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -5 \, a - b + 2 \, c \\ a \\ -3 \, b + 2 \, c \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -5 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 0 \\ -3 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 17 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} & & & & x_{3} &-& x_{4} &=& 0 \\-x_{1} &-& 2 \, x_{2} &+& 6 \, x_{3} &-& 3 \, x_{4} &=& 0 \\-3 \, x_{1} &-& 6 \, x_{2} &+& 10 \, x_{3} &-& x_{4} &=& 0 \\-2 \, x_{1} &-& 4 \, x_{2} &+& 12 \, x_{3} &-& 6 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 0 & 0 & 1 & -1 & 0 \\ -1 & -2 & 6 & -3 & 0 \\ -3 & -6 & 10 & -1 & 0 \\ -2 & -4 & 12 & -6 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 2 & 0 & -3 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -2 \, a + 3 \, b \\ a \\ b \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 18 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} -x_{1} &-& 4 \, x_{2} &-& 7 \, x_{3} &-& 6 \, x_{4} &=& 0 \\x_{1} &+& 4 \, x_{2} &+& 6 \, x_{3} &+& 5 \, x_{4} &=& 0 \\-2 \, x_{1} &-& 8 \, x_{2} &-& 12 \, x_{3} &-& 10 \, x_{4} &=& 0 \\3 \, x_{1} &+& 12 \, x_{2} &+& 10 \, x_{3} &+& 7 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} -1 & -4 & -7 & -6 & 0 \\ 1 & 4 & 6 & 5 & 0 \\ -2 & -8 & -12 & -10 & 0 \\ 3 & 12 & 10 & 7 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 4 & 0 & -1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -4 \, a + b \\ a \\ -b \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 0 \\ -1 \\ 1 \end{array}\right] \right\} \).

Example 19 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} & & &+& 2 \, x_{3} &+& 2 \, x_{4} &=& 0 \\ & & x_{2} &+& 2 \, x_{3} &-& x_{4} &=& 0 \\ & & 5 \, x_{2} &+& 10 \, x_{3} &-& 5 \, x_{4} &=& 0 \\x_{1} &-& 7 \, x_{2} &-& 12 \, x_{3} &+& 9 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & 0 & 2 & 2 & 0 \\ 0 & 1 & 2 & -1 & 0 \\ 0 & 5 & 10 & -5 & 0 \\ 1 & -7 & -12 & 9 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 2 & 2 & 0 \\ 0 & 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -2 \, a - 2 \, b \\ -2 \, a + b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -2 \\ -2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 20 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &+& x_{2} &+& 2 \, x_{3} &+& x_{4} &-& x_{5} &=& 0 \\-x_{1} &-& 3 \, x_{2} &-& 4 \, x_{3} &-& 7 \, x_{4} &-& x_{5} &=& 0 \\ & & x_{2} &+& x_{3} &+& 3 \, x_{4} &+& x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a + 2 \, b + 2 \, c \\ -a - 3 \, b - c \\ a \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -3 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ -1 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 21 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &-& 3 \, x_{2} &-& 11 \, x_{3} &=& 0 \\x_{1} &-& 2 \, x_{2} &-& 8 \, x_{3} &=& 0 \\ &-& 4 \, x_{2} &-& 12 \, x_{3} &=& 0 \\ &-& 2 \, x_{2} &-& 6 \, x_{3} &=& 0 \\ &-& 3 \, x_{2} &-& 9 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 1 & -3 & -11 & 0 \\ 1 & -2 & -8 & 0 \\ 0 & -4 & -12 & 0 \\ 0 & -2 & -6 & 0 \\ 0 & -3 & -9 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -2 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 2 \, a \\ -3 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 2 \\ -3 \\ 1 \end{array}\right] \right\} \).

Example 22 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} 2 \, x_{1} &+& 2 \, x_{2} & & &=& 0 \\-4 \, x_{1} &-& 11 \, x_{2} & & &=& 0 \\-4 \, x_{1} &-& 9 \, x_{2} & & &=& 0 \\ & & 4 \, x_{2} & & &=& 0 \\-x_{1} &+& x_{2} & & &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 2 & 2 & 0 & 0 \\ -4 & -11 & 0 & 0 \\ -4 & -9 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ -1 & 1 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 23 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &-& x_{2} &+& 2 \, x_{3} &=& 0 \\-2 \, x_{1} &+& 3 \, x_{2} &-& 5 \, x_{3} &=& 0 \\ &-& 2 \, x_{2} &+& 2 \, x_{3} &=& 0 \\5 \, x_{1} &-& 6 \, x_{2} &+& 11 \, x_{3} &=& 0 \\3 \, x_{1} &-& 7 \, x_{2} &+& 10 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 1 & -1 & 2 & 0 \\ -2 & 3 & -5 & 0 \\ 0 & -2 & 2 & 0 \\ 5 & -6 & 11 & 0 \\ 3 & -7 & 10 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 24 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &-& 5 \, x_{2} &+& 10 \, x_{3} &=& 0 \\ & & x_{2} &-& 2 \, x_{3} &=& 0 \\ & & x_{2} &-& 2 \, x_{3} &=& 0 \\ & & x_{2} &-& 2 \, x_{3} &=& 0 \\x_{1} &-& 6 \, x_{2} &+& 12 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 1 & -5 & 10 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 1 & -2 & 0 \\ 1 & -6 & 12 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 25 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &-& x_{2} &+& 5 \, x_{3} &-& 8 \, x_{4} &=& 0 \\2 \, x_{1} &-& x_{2} &+& 5 \, x_{3} &-& 8 \, x_{4} &=& 0 \\ &-& x_{2} &+& 6 \, x_{3} &-& 10 \, x_{4} &=& 0 \\ & & 2 \, x_{2} &-& 7 \, x_{3} &+& 10 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & -1 & 5 & -8 & 0 \\ 2 & -1 & 5 & -8 & 0 \\ 0 & -1 & 6 & -10 & 0 \\ 0 & 2 & -7 & 10 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 26 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} 4 \, x_{1} &-& 7 \, x_{2} &+& 11 \, x_{3} &=& 0 \\x_{1} &-& 3 \, x_{2} &+& 4 \, x_{3} &=& 0 \\ & & 4 \, x_{2} &-& 4 \, x_{3} &=& 0 \\-x_{1} &-& x_{2} & & &=& 0 \\-x_{1} &+& x_{2} &-& 2 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 4 & -7 & 11 & 0 \\ 1 & -3 & 4 & 0 \\ 0 & 4 & -4 & 0 \\ -1 & -1 & 0 & 0 \\ -1 & 1 & -2 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 27 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} -2 \, x_{1} &+& 6 \, x_{2} &+& 7 \, x_{3} &=& 0 \\3 \, x_{1} &-& 9 \, x_{2} &-& 11 \, x_{3} &=& 0 \\-2 \, x_{1} &+& 6 \, x_{2} &+& 3 \, x_{3} &=& 0 \\-2 \, x_{1} &+& 6 \, x_{2} &+& 8 \, x_{3} &=& 0 \\-2 \, x_{1} &+& 6 \, x_{2} &+& 3 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} -2 & 6 & 7 & 0 \\ 3 & -9 & -11 & 0 \\ -2 & 6 & 3 & 0 \\ -2 & 6 & 8 & 0 \\ -2 & 6 & 3 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 3 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 3 \\ 1 \\ 0 \end{array}\right] \right\} \).

Example 28 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} -x_{1} &-& 3 \, x_{2} &+& 4 \, x_{3} &=& 0 \\ & & x_{2} &-& x_{3} &=& 0 \\x_{1} &-& x_{2} & & &=& 0 \\-x_{1} &+& 6 \, x_{2} &-& 5 \, x_{3} &=& 0 \\-2 \, x_{1} &+& 11 \, x_{2} &-& 9 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} -1 & -3 & 4 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & -1 & 0 & 0 \\ -1 & 6 & -5 & 0 \\ -2 & 11 & -9 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \).

Example 29 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &-& 8 \, x_{2} &+& 4 \, x_{3} &+& 4 \, x_{4} &+& 9 \, x_{5} &=& 0 \\ & & x_{2} & & &-& x_{4} &-& x_{5} &=& 0 \\ & & 3 \, x_{2} &+& x_{3} &-& 4 \, x_{4} &-& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & -8 & 4 & 4 & 9 & 0 \\ 0 & 1 & 0 & -1 & -1 & 0 \\ 0 & 3 & 1 & -4 & -3 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & -1 & -1 & 0 \\ 0 & 0 & 1 & -1 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -b \\ a + b \\ a \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 30 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &+& x_{2} &-& 3 \, x_{3} &+& 7 \, x_{4} &=& 0 \\ & & x_{2} &-& 4 \, x_{3} &+& 7 \, x_{4} &=& 0 \\2 \, x_{1} &+& x_{2} &-& x_{3} &+& 5 \, x_{4} &=& 0 \\ &-& x_{2} &-& x_{3} &+& 3 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} -2 \, a \\ a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 31 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} -x_{1} &+& 4 \, x_{2} &-& 10 \, x_{3} &-& 6 \, x_{4} &+& 8 \, x_{5} &=& 0 \\-2 \, x_{1} &+& x_{2} &+& x_{3} &+& 2 \, x_{4} &+& 2 \, x_{5} &=& 0 \\x_{1} &-& 2 \, x_{2} &+& 4 \, x_{3} &+& 2 \, x_{4} &-& 4 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} -1 & 4 & -10 & -6 & 8 & 0 \\ -2 & 1 & 1 & 2 & 2 & 0 \\ 1 & -2 & 4 & 2 & -4 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & -2 & -2 & 0 & 0 \\ 0 & 1 & -3 & -2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 2 \, a + 2 \, b \\ 3 \, a + 2 \, b - 2 \, c \\ a \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 2 \\ 3 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 2 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ -2 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 32 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &-& 8 \, x_{2} &+& x_{3} &+& 7 \, x_{4} &=& 0 \\5 \, x_{1} &-& 9 \, x_{2} &+& 5 \, x_{3} &+& 4 \, x_{4} &=& 0 \\ & & 2 \, x_{2} & & &-& 2 \, x_{4} &=& 0 \\3 \, x_{1} &-& x_{2} &+& 3 \, x_{3} &-& 2 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & -8 & 1 & 7 & 0 \\ 5 & -9 & 5 & 4 & 0 \\ 0 & 2 & 0 & -2 & 0 \\ 3 & -1 & 3 & -2 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 1 & -1 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a + b \\ b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 33 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} & & &-& 3 \, x_{3} &-& 8 \, x_{4} &-& 2 \, x_{5} &=& 0 \\ & & x_{2} &+& x_{3} &+& 2 \, x_{4} &+& 4 \, x_{5} &=& 0 \\ &-& x_{2} & & & & &-& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} 2 \, a - b \\ -3 \, b \\ -2 \, a - b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ -2 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -1 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 34 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &+& 2 \, x_{2} &-& 3 \, x_{3} &=& 0 \\2 \, x_{1} &+& 11 \, x_{2} &-& 6 \, x_{3} &=& 0 \\x_{1} &+& 3 \, x_{2} &-& 3 \, x_{3} &=& 0 \\ & & 3 \, x_{2} & & &=& 0 \\3 \, x_{1} &+& 11 \, x_{2} &-& 9 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 1 & 2 & -3 & 0 \\ 2 & 11 & -6 & 0 \\ 1 & 3 & -3 & 0 \\ 0 & 3 & 0 & 0 \\ 3 & 11 & -9 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -3 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 3 \, a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 3 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 35 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &+& 2 \, x_{2} &+& 11 \, x_{3} &-& 11 \, x_{4} &-& 12 \, x_{5} &=& 0 \\-x_{1} &-& x_{2} &-& 8 \, x_{3} &+& 8 \, x_{4} &+& 8 \, x_{5} &=& 0 \\x_{1} &+& 2 \, x_{2} &+& 11 \, x_{3} &-& 10 \, x_{4} &-& 11 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & 2 & 11 & -11 & -12 & 0 \\ -1 & -1 & -8 & 8 & 8 & 0 \\ 1 & 2 & 11 & -10 & -11 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 5 & 0 & 1 & 0 \\ 0 & 1 & 3 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -5 \, a - b \\ -3 \, a + b \\ a \\ -b \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -5 \\ -3 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ -1 \\ 1 \end{array}\right] \right\} \).

Example 36 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} -2 \, x_{1} &-& 5 \, x_{2} &-& 8 \, x_{3} &=& 0 \\x_{1} &-& 4 \, x_{2} &-& 9 \, x_{3} &=& 0 \\x_{1} & & &-& x_{3} &=& 0 \\-x_{1} &+& 2 \, x_{2} &+& 5 \, x_{3} &=& 0 \\-x_{1} & & &+& x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} -2 & -5 & -8 & 0 \\ 1 & -4 & -9 & 0 \\ 1 & 0 & -1 & 0 \\ -1 & 2 & 5 & 0 \\ -1 & 0 & 1 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} a \\ -2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right] \right\} \).

Example 37 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} -3 \, x_{1} &+& 4 \, x_{2} &-& 2 \, x_{3} &-& 8 \, x_{4} &+& 3 \, x_{5} &=& 0 \\x_{1} & & &+& 3 \, x_{3} &-& x_{4} &+& 11 \, x_{5} &=& 0 \\4 \, x_{1} &-& 5 \, x_{2} &+& 3 \, x_{3} &+& 10 \, x_{4} &-& 2 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} -3 & 4 & -2 & -8 & 3 & 0 \\ 1 & 0 & 3 & -1 & 11 & 0 \\ 4 & -5 & 3 & 10 & -2 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 2 & -1 & 0 \\ 0 & 1 & 0 & -1 & 2 & 0 \\ 0 & 0 & 1 & -1 & 4 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -2 \, a + b \\ a - 2 \, b \\ a - 4 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -2 \\ 1 \\ 1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ -2 \\ -4 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 38 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} -x_{1} &-& 5 \, x_{2} &-& x_{3} &+& 7 \, x_{4} &=& 0 \\x_{1} &+& 5 \, x_{2} &-& 2 \, x_{3} &+& 5 \, x_{4} &=& 0 \\-x_{1} &-& 5 \, x_{2} &+& x_{3} &-& x_{4} &=& 0 \\ & & &-& 3 \, x_{3} &+& 12 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} -1 & -5 & -1 & 7 & 0 \\ 1 & 5 & -2 & 5 & 0 \\ -1 & -5 & 1 & -1 & 0 \\ 0 & 0 & -3 & 12 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 5 & 0 & -3 & 0 \\ 0 & 0 & 1 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -5 \, a + 3 \, b \\ a \\ 4 \, b \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -5 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 4 \\ 1 \end{array}\right] \right\} \).

Example 39 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &-& 2 \, x_{2} &+& 5 \, x_{3} &+& x_{4} & & &=& 0 \\x_{1} &-& x_{2} &+& 3 \, x_{3} & & &+& x_{5} &=& 0 \\ & & 3 \, x_{2} &-& 6 \, x_{3} &-& 3 \, x_{4} &+& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a + b - 2 \, c \\ 2 \, a + b - c \\ a \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ -1 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 40 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} 3 \, x_{1} &+& 10 \, x_{2} &-& 6 \, x_{3} &=& 0 \\-x_{1} &-& 3 \, x_{2} &+& 2 \, x_{3} &=& 0 \\x_{1} &-& x_{2} &-& 2 \, x_{3} &=& 0 \\3 \, x_{1} &+& 10 \, x_{2} &-& 6 \, x_{3} &=& 0 \\ & & 5 \, x_{2} & & &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 3 & 10 & -6 & 0 \\ -1 & -3 & 2 & 0 \\ 1 & -1 & -2 & 0 \\ 3 & 10 & -6 & 0 \\ 0 & 5 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 2 \, a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 41 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} x_{1} &-& 4 \, x_{2} &-& 6 \, x_{3} &=& 0 \\x_{1} &-& 3 \, x_{2} &-& 5 \, x_{3} &=& 0 \\ & & & & 0 &=& 0 \\ &-& 5 \, x_{2} &-& 5 \, x_{3} &=& 0 \\x_{1} &-& x_{2} &-& 3 \, x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} 2 \, a \\ -a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 2 \\ -1 \\ 1 \end{array}\right] \right\} \).

Example 42 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} x_{1} &+& x_{2} &+& x_{3} &-& 2 \, x_{4} &=& 0 \\-x_{1} & & &-& 4 \, x_{3} &+& 4 \, x_{4} &=& 0 \\ & & x_{2} &-& 3 \, x_{3} &+& 3 \, x_{4} &=& 0 \\5 \, x_{1} &+& 3 \, x_{2} &+& 11 \, x_{3} &-& 12 \, x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{cccc|c} 1 & 1 & 1 & -2 & 0 \\ -1 & 0 & -4 & 4 & 0 \\ 0 & 1 & -3 & 3 & 0 \\ 5 & 3 & 11 & -12 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 4 & 0 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -4 \, a \\ 3 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -4 \\ 3 \\ 1 \\ 0 \end{array}\right] \right\} \).

Example 43 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} & & &-& 5 \, x_{3} &+& 9 \, x_{4} &-& 9 \, x_{5} &=& 0 \\ & & x_{2} &-& 5 \, x_{3} &+& 3 \, x_{4} &-& 11 \, x_{5} &=& 0 \\-x_{1} & & &+& 6 \, x_{3} &-& 10 \, x_{4} &+& 11 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & 0 & -5 & 9 & -9 & 0 \\ 0 & 1 & -5 & 3 & -11 & 0 \\ -1 & 0 & 6 & -10 & 11 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 4 & 1 & 0 \\ 0 & 1 & 0 & -2 & -1 & 0 \\ 0 & 0 & 1 & -1 & 2 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -4 \, a - b \\ 2 \, a + b \\ a - 2 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -4 \\ 2 \\ 1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ -2 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 44 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &-& x_{2} &-& 6 \, x_{3} &+& 4 \, x_{4} & & &=& 0 \\ & & x_{2} &+& 2 \, x_{3} &-& 2 \, x_{4} &-& x_{5} &=& 0 \\-2 \, x_{1} &-& x_{2} &+& 6 \, x_{3} &-& 2 \, x_{4} &+& 3 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} 4 \, a - 2 \, b + c \\ -2 \, a + 2 \, b + c \\ a \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 4 \\ -2 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ 2 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 45 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} & & 5 \, x_{2} &-& 10 \, x_{3} &=& 0 \\x_{1} &+& 6 \, x_{2} &-& 11 \, x_{3} &=& 0 \\ &-& 4 \, x_{2} &+& 8 \, x_{3} &=& 0 \\x_{1} &+& x_{2} &-& x_{3} &=& 0 \\x_{1} & & &+& x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 0 & 5 & -10 & 0 \\ 1 & 6 & -11 & 0 \\ 0 & -4 & 8 & 0 \\ 1 & 1 & -1 & 0 \\ 1 & 0 & 1 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 2 \\ 1 \end{array}\right] \right\} \).

Example 46 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} & & x_{2} & & &+& 4 \, x_{4} &+& 3 \, x_{5} &=& 0 \\-x_{1} &-& x_{2} &-& 2 \, x_{3} &-& x_{4} &-& 8 \, x_{5} &=& 0 \\-2 \, x_{1} &-& x_{2} &-& 3 \, x_{3} &+& 2 \, x_{4} &-& 11 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 0 & 1 & 0 & 4 & 3 & 0 \\ -1 & -1 & -2 & -1 & -8 & 0 \\ -2 & -1 & -3 & 2 & -11 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & -3 & 1 & 0 \\ 0 & 1 & 0 & 4 & 3 & 0 \\ 0 & 0 & 1 & 0 & 2 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} 3 \, a - b \\ -4 \, a - 3 \, b \\ -2 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 3 \\ -4 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -2 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 47 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{5} -2 \, x_{1} & & & & &+& 2 \, x_{4} &=& 0 \\-x_{1} &-& x_{2} & & &-& 2 \, x_{4} &=& 0 \\3 \, x_{1} &+& 2 \, x_{2} & & &+& 3 \, x_{4} &=& 0 \\-x_{1} & & & & &+& x_{4} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} b \\ -3 \, b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 1 \\ -3 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 48 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} & & &+& 2 \, x_{3} &-& x_{4} &+& x_{5} &=& 0 \\ & & x_{2} &-& 3 \, x_{3} &+& 5 \, x_{4} & & &=& 0 \\-4 \, x_{1} &-& 2 \, x_{2} &-& x_{3} &-& 7 \, x_{4} &-& 5 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

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

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a - 3 \, b \\ -2 \, a + 3 \, b \\ a + b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ -2 \\ 1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -3 \\ 3 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 49 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{6} x_{1} &+& x_{2} &-& 9 \, x_{3} &+& 10 \, x_{4} &+& 9 \, x_{5} &=& 0 \\ & & x_{2} &-& 5 \, x_{3} &+& 6 \, x_{4} &+& 4 \, x_{5} &=& 0 \\x_{1} & & &-& 3 \, x_{3} &+& 3 \, x_{4} &+& 4 \, x_{5} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & 1 & -9 & 10 & 9 & 0 \\ 0 & 1 & -5 & 6 & 4 & 0 \\ 1 & 0 & -3 & 3 & 4 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -b \\ -a + b \\ a + b \\ a \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} 0 \\ -1 \\ 1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 1 \\ 0 \\ 1 \end{array}\right] \right\} \).

Example 50 πŸ”—

Consider the homogeneous system of equations \begin{alignat*}{4} 4 \, x_{1} &-& 12 \, x_{2} &+& 4 \, x_{3} &=& 0 \\3 \, x_{1} &-& 5 \, x_{2} &+& 3 \, x_{3} &=& 0 \\x_{1} &-& 5 \, x_{2} &+& x_{3} &=& 0 \\3 \, x_{1} &-& 10 \, x_{2} &+& 3 \, x_{3} &=& 0 \\x_{1} &-& 2 \, x_{2} &+& x_{3} &=& 0 \\ \end{alignat*}

  1. Find the solution space of this system.
  2. Find a basis of the solution space.

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc|c} 4 & -12 & 4 & 0 \\ 3 & -5 & 3 & 0 \\ 1 & -5 & 1 & 0 \\ 3 & -10 & 3 & 0 \\ 1 & -2 & 1 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

  1. The solution space is \( \left\{ \left[\begin{array}{c} -a \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \( \left\{ \left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right] \right\} \).