V10 - Basis of Solution Space


Example 1

V10 - Basis of Solution Space (ver. 1)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + x_{2} &=0\\ x_{2} + x_{3} &=0\\ -4 \, x_{1} + 12 \, x_{2} + 8 \, x_{3} &=0\\ -2 \, x_{1} + 3 \, x_{2} + x_{3} &=0\\ -2 \, x_{1} + 5 \, x_{2} + 3 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & 1 & 1 \\ -4 & 12 & 8 \\ -2 & 3 & 1 \\ -2 & 5 & 3 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 2

V10 - Basis of Solution Space (ver. 2)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 5 \, x_{2} + 7 \, x_{3} &=0\\ x_{2} + 2 \, x_{3} &=0\\ 4 \, x_{2} + 8 \, x_{3} &=0\\ 2 \, x_{1} + 8 \, x_{2} + 10 \, x_{3} &=0\\ -2 \, x_{1} - 7 \, x_{2} - 8 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 1 & 5 & 7 \\ 0 & 1 & 2 \\ 0 & 4 & 8 \\ 2 & 8 & 10 \\ -2 & -7 & -8 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 3 \\ -2 \\ 1 \end{array}\right]\right\}.\]


Example 3

V10 - Basis of Solution Space (ver. 3)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + x_{2} + 2 \, x_{4} &=0\\ -3 \, x_{1} - 2 \, x_{2} - x_{3} - 7 \, x_{4} &=0\\ x_{3} + 2 \, x_{4} &=0\\ -2 \, x_{2} - 5 \, x_{3} - 12 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & 1 & 0 & 2 \\ -3 & -2 & -1 & -7 \\ 0 & 0 & 1 & 2 \\ 0 & -2 & -5 & -12 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -1 \\ -2 \\ 1 \end{array}\right]\right\}.\]


Example 4

V10 - Basis of Solution Space (ver. 4)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{2} + 2 \, x_{4} &=0\\ x_{1} + x_{3} - 2 \, x_{4} &=0\\ 2 \, x_{1} + 2 \, x_{3} - 4 \, x_{4} &=0\\ -3 \, x_{1} + 7 \, x_{2} - 3 \, x_{3} - 8 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 0 & -1 & 0 & 2 \\ 1 & 0 & 1 & -2 \\ 2 & 0 & 2 & -4 \\ -3 & 7 & -3 & -8 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 1 & -2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ 2 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 5

V10 - Basis of Solution Space (ver. 5)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - x_{2} - 2 \, x_{3} + 5 \, x_{4} - 7 \, x_{5} &=0\\ x_{2} - 4 \, x_{3} - 4 \, x_{4} - 7 \, x_{5} &=0\\ x_{2} - 3 \, x_{3} - 4 \, x_{4} - 5 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & -1 & -2 & 5 & -7 \\ 0 & 1 & -4 & -4 & -7 \\ 0 & 1 & -3 & -4 & -5 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 0 & -4 & 1 \\ 0 & 0 & 1 & 0 & 2 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 4 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ -1 \\ -2 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 6

V10 - Basis of Solution Space (ver. 6)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 4 \, x_{2} + x_{3} &=0\\ -5 \, x_{1} - 9 \, x_{2} - 5 \, x_{3} &=0\\ -2 \, x_{1} - 2 \, x_{3} &=0\\ x_{1} - 3 \, x_{2} + x_{3} &=0\\ 2 \, x_{2} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 1 & 4 & 1 \\ -5 & -9 & -5 \\ -2 & 0 & -2 \\ 1 & -3 & 1 \\ 0 & 2 & 0 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 7

V10 - Basis of Solution Space (ver. 7)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} - x_{2} - 4 \, x_{3} &=0\\ -x_{1} - 2 \, x_{2} - 7 \, x_{3} &=0\\ -x_{1} - 2 \, x_{2} - 7 \, x_{3} &=0\\ -2 \, x_{1} - 2 \, x_{2} - 8 \, x_{3} &=0\\ x_{1} + x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} -1 & -1 & -4 \\ -1 & -2 & -7 \\ -1 & -2 & -7 \\ -2 & -2 & -8 \\ 1 & 0 & 1 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -3 \\ 1 \end{array}\right]\right\}.\]


Example 8

V10 - Basis of Solution Space (ver. 8)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -2 \, x_{1} - 10 \, x_{2} + 2 \, x_{3} &=0\\ 2 \, x_{1} + 10 \, x_{2} + 11 \, x_{3} &=0\\ -2 \, x_{1} - 10 \, x_{2} - 9 \, x_{3} &=0\\ -2 \, x_{1} - 10 \, x_{2} - 9 \, x_{3} &=0\\ x_{1} + 5 \, x_{2} + x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} -2 & -10 & 2 \\ 2 & 10 & 11 \\ -2 & -10 & -9 \\ -2 & -10 & -9 \\ 1 & 5 & 1 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 5 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -5 \\ 1 \\ 0 \end{array}\right]\right\}.\]


Example 9

V10 - Basis of Solution Space (ver. 9)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + 5 \, x_{2} - 3 \, x_{3} - 4 \, x_{4} &=0\\ x_{3} + x_{4} &=0\\ x_{1} - 5 \, x_{2} + 5 \, x_{3} + 6 \, x_{4} &=0\\ -2 \, x_{1} + 10 \, x_{2} - 6 \, x_{3} - 8 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -1 & 5 & -3 & -4 \\ 0 & 0 & 1 & 1 \\ 1 & -5 & 5 & 6 \\ -2 & 10 & -6 & -8 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & -5 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 5 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ 0 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 10

V10 - Basis of Solution Space (ver. 10)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + x_{2} - 4 \, x_{3} &=0\\ x_{1} - 2 \, x_{2} + 7 \, x_{3} &=0\\ -x_{1} + 3 \, x_{2} - 10 \, x_{3} &=0\\ x_{1} - x_{2} + 4 \, x_{3} &=0\\ 0 &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} -1 & 1 & -4 \\ 1 & -2 & 7 \\ -1 & 3 & -10 \\ 1 & -1 & 4 \\ 0 & 0 & 0 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 3 \\ 1 \end{array}\right]\right\}.\]


Example 11

V10 - Basis of Solution Space (ver. 11)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + 5 \, x_{2} - 8 \, x_{3} + 11 \, x_{4} &=0\\ x_{3} - x_{4} &=0\\ -x_{1} + 5 \, x_{2} - 3 \, x_{3} + 6 \, x_{4} &=0\\ -x_{1} + 5 \, x_{2} + 4 \, x_{3} - x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -1 & 5 & -8 & 11 \\ 0 & 0 & 1 & -1 \\ -1 & 5 & -3 & 6 \\ -1 & 5 & 4 & -1 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & -5 & 0 & -3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 5 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} 3 \\ 0 \\ 1 \\ 1 \end{array}\right]\right\}.\]


Example 12

V10 - Basis of Solution Space (ver. 12)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - 2 \, x_{2} - 6 \, x_{3} + 7 \, x_{4} &=0\\ x_{2} + 3 \, x_{3} - 4 \, x_{4} &=0\\ 2 \, x_{2} + 7 \, x_{3} - 10 \, x_{4} &=0\\ -3 \, x_{2} - 5 \, x_{3} + 4 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & -2 & -6 & 7 \\ 0 & 1 & 3 & -4 \\ 0 & 2 & 7 & -10 \\ 0 & -3 & -5 & 4 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ -2 \\ 2 \\ 1 \end{array}\right]\right\}.\]


Example 13

V10 - Basis of Solution Space (ver. 13)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 2 \, x_{3} - 4 \, x_{4} - 7 \, x_{5} &=0\\ -3 \, x_{1} + x_{2} - 3 \, x_{3} + 9 \, x_{4} + 10 \, x_{5} &=0\\ x_{3} - x_{4} - 4 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 0 & 2 & -4 & -7 \\ -3 & 1 & -3 & 9 & 10 \\ 0 & 0 & 1 & -1 & -4 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & -2 & 1 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & -4 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ 0 \\ 1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ -1 \\ 4 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 14

V10 - Basis of Solution Space (ver. 14)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + 4 \, x_{2} + 7 \, x_{3} + 7 \, x_{4} + 4 \, x_{5} &=0\\ -x_{1} + 3 \, x_{2} + 6 \, x_{3} + 5 \, x_{4} + 3 \, x_{5} &=0\\ -2 \, x_{1} + 2 \, x_{2} + 8 \, x_{3} + 2 \, x_{4} + 2 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} -1 & 4 & 7 & 7 & 4 \\ -1 & 3 & 6 & 5 & 3 \\ -2 & 2 & 8 & 2 & 2 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & -3 & 1 & 0 \\ 0 & 1 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 3 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ -2 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 0 \\ -1 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 15

V10 - Basis of Solution Space (ver. 15)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 2 \, x_{2} - 7 \, x_{3} &=0\\ -2 \, x_{1} - 3 \, x_{2} + 10 \, x_{3} &=0\\ x_{1} - x_{2} + 5 \, x_{3} &=0\\ -x_{1} - x_{3} &=0\\ 2 \, x_{1} + 2 \, x_{2} - 6 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 1 & 2 & -7 \\ -2 & -3 & 10 \\ 1 & -1 & 5 \\ -1 & 0 & -1 \\ 2 & 2 & -6 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 4 \\ 1 \end{array}\right]\right\}.\]


Example 16

V10 - Basis of Solution Space (ver. 16)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - 5 \, x_{2} - 9 \, x_{3} + 11 \, x_{4} &=0\\ x_{1} - 4 \, x_{2} - 8 \, x_{3} + 10 \, x_{4} &=0\\ x_{1} - 3 \, x_{3} + 4 \, x_{4} &=0\\ -x_{1} + 5 \, x_{2} + 6 \, x_{3} - 5 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & -5 & -9 & 11 \\ 1 & -4 & -8 & 10 \\ 1 & 0 & -3 & 4 \\ -1 & 5 & 6 & -5 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ -1 \\ 2 \\ 1 \end{array}\right]\right\}.\]


Example 17

V10 - Basis of Solution Space (ver. 17)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - 3 \, x_{2} - 5 \, x_{3} + 10 \, x_{4} &=0\\ x_{1} - 2 \, x_{2} - 4 \, x_{3} + 7 \, x_{4} &=0\\ x_{1} + x_{2} - x_{3} - 2 \, x_{4} &=0\\ -2 \, x_{2} - 2 \, x_{3} + 6 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & -3 & -5 & 10 \\ 1 & -2 & -4 & 7 \\ 1 & 1 & -1 & -2 \\ 0 & -2 & -2 & 6 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ 3 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 18

V10 - Basis of Solution Space (ver. 18)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} 4 \, x_{1} - 4 \, x_{2} + 8 \, x_{3} &=0\\ -x_{1} - x_{2} &=0\\ x_{1} - 2 \, x_{2} + 3 \, x_{3} &=0\\ -x_{1} - 6 \, x_{2} + 5 \, x_{3} &=0\\ 4 \, x_{2} - 4 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 4 & -4 & 8 \\ -1 & -1 & 0 \\ 1 & -2 & 3 \\ -1 & -6 & 5 \\ 0 & 4 & -4 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 1 \\ 1 \end{array}\right]\right\}.\]


Example 19

V10 - Basis of Solution Space (ver. 19)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - 2 \, x_{2} - x_{3} - 2 \, x_{4} &=0\\ x_{2} - x_{4} &=0\\ x_{1} + 8 \, x_{2} - 9 \, x_{4} &=0\\ 3 \, x_{2} - 3 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & -2 & -1 & -2 \\ 0 & 1 & 0 & -1 \\ 1 & 8 & 0 & -9 \\ 0 & 3 & 0 & -3 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ 1 \\ -3 \\ 1 \end{array}\right]\right\}.\]


Example 20

V10 - Basis of Solution Space (ver. 20)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 2 \, x_{2} + 6 \, x_{3} - 12 \, x_{4} &=0\\ x_{2} + x_{3} - 3 \, x_{4} &=0\\ 3 \, x_{2} + 4 \, x_{3} - 11 \, x_{4} &=0\\ x_{1} - 2 \, x_{2} + 6 \, x_{3} - 8 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & 2 & 6 & -12 \\ 0 & 1 & 1 & -3 \\ 0 & 3 & 4 & -11 \\ 1 & -2 & 6 & -8 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -2 \\ 1 \\ 2 \\ 1 \end{array}\right]\right\}.\]


Example 21

V10 - Basis of Solution Space (ver. 21)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 3 \, x_{2} - x_{3} - 6 \, x_{4} - 2 \, x_{5} &=0\\ x_{2} - x_{3} - x_{4} &=0\\ -x_{1} - x_{2} - x_{3} + 4 \, x_{4} + 2 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 3 & -1 & -6 & -2 \\ 0 & 1 & -1 & -1 & 0 \\ -1 & -1 & -1 & 4 & 2 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 2 & -3 & -2 \\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -2 \\ 1 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} 3 \\ 1 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 22

V10 - Basis of Solution Space (ver. 22)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} - x_{2} - 2 \, x_{3} - 3 \, x_{4} + 2 \, x_{5} &=0\\ -x_{1} - 2 \, x_{2} - 3 \, x_{3} - 5 \, x_{4} + 3 \, x_{5} &=0\\ 2 \, x_{1} + 3 \, x_{2} + 5 \, x_{3} + 8 \, x_{4} - 5 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} -1 & -1 & -2 & -3 & 2 \\ -1 & -2 & -3 & -5 & 3 \\ 2 & 3 & 5 & 8 & -5 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 1 & 1 & -1 \\ 0 & 1 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ -2 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 1 \\ 1 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 23

V10 - Basis of Solution Space (ver. 23)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} - 2 \, x_{2} + x_{3} - 4 \, x_{4} - 3 \, x_{5} &=0\\ x_{3} - 3 \, x_{4} - 3 \, x_{5} &=0\\ -x_{1} - 2 \, x_{2} + 2 \, x_{3} - 7 \, x_{4} - 6 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} -1 & -2 & 1 & -4 & -3 \\ 0 & 0 & 1 & -3 & -3 \\ -1 & -2 & 2 & -7 & -6 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 2 & 0 & 1 & 0 \\ 0 & 0 & 1 & -3 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ 0 \\ 3 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 0 \\ 0 \\ 3 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 24

V10 - Basis of Solution Space (ver. 24)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + x_{4} &=0\\ -4 \, x_{1} + x_{2} - 5 \, x_{3} - 6 \, x_{4} &=0\\ -2 \, x_{1} - 2 \, x_{2} + 11 \, x_{3} + 2 \, x_{4} &=0\\ 5 \, x_{1} + 5 \, x_{3} + 5 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ -4 & 1 & -5 & -6 \\ -2 & -2 & 11 & 2 \\ 5 & 0 & 5 & 5 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 2 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 25

V10 - Basis of Solution Space (ver. 25)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -2 \, x_{1} + 5 \, x_{3} + 6 \, x_{4} &=0\\ -x_{1} + 2 \, x_{2} - 5 \, x_{3} + 5 \, x_{4} &=0\\ 3 \, x_{1} - x_{2} - 4 \, x_{3} - 10 \, x_{4} &=0\\ -2 \, x_{1} + 6 \, x_{3} + 6 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -2 & 0 & 5 & 6 \\ -1 & 2 & -5 & 5 \\ 3 & -1 & -4 & -10 \\ -2 & 0 & 6 & 6 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 3 \\ -1 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 26

V10 - Basis of Solution Space (ver. 26)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} 2 \, x_{1} + 7 \, x_{2} + 9 \, x_{3} &=0\\ x_{1} + 5 \, x_{2} + 6 \, x_{3} &=0\\ 4 \, x_{2} + 4 \, x_{3} &=0\\ x_{1} + 3 \, x_{2} + 4 \, x_{3} &=0\\ -x_{1} - 4 \, x_{2} - 5 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 2 & 7 & 9 \\ 1 & 5 & 6 \\ 0 & 4 & 4 \\ 1 & 3 & 4 \\ -1 & -4 & -5 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 27

V10 - Basis of Solution Space (ver. 27)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 2 \, x_{2} + 5 \, x_{3} - 9 \, x_{4} - x_{5} &=0\\ x_{2} + 3 \, x_{3} - 5 \, x_{4} - x_{5} &=0\\ x_{2} + 4 \, x_{3} - 7 \, x_{4} - x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 2 & 5 & -9 & -1 \\ 0 & 1 & 3 & -5 & -1 \\ 0 & 1 & 4 & -7 & -1 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & -1 & 1 \\ 0 & 1 & 0 & 1 & -1 \\ 0 & 0 & 1 & -2 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ -1 \\ 2 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 28

V10 - Basis of Solution Space (ver. 28)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + 5 \, x_{2} - 8 \, x_{3} - 2 \, x_{4} &=0\\ -x_{1} - 4 \, x_{2} + 3 \, x_{3} &=0\\ x_{1} + 2 \, x_{3} + x_{4} &=0\\ 2 \, x_{1} + 7 \, x_{2} - 10 \, x_{3} - 5 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -1 & 5 & -8 & -2 \\ -1 & -4 & 3 & 0 \\ 1 & 0 & 2 & 1 \\ 2 & 7 & -10 & -5 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 29

V10 - Basis of Solution Space (ver. 29)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -3 \, x_{1} + 3 \, x_{2} - 3 \, x_{3} &=0\\ x_{1} - 2 \, x_{2} + x_{3} &=0\\ 4 \, x_{1} - 11 \, x_{2} + 4 \, x_{3} &=0\\ -4 \, x_{1} + 8 \, x_{2} - 4 \, x_{3} &=0\\ 2 \, x_{1} - 3 \, x_{2} + 2 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} -3 & 3 & -3 \\ 1 & -2 & 1 \\ 4 & -11 & 4 \\ -4 & 8 & -4 \\ 2 & -3 & 2 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 30

V10 - Basis of Solution Space (ver. 30)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -2 \, x_{1} + 7 \, x_{2} + 8 \, x_{3} - x_{4} &=0\\ -3 \, x_{1} + 10 \, x_{2} + 12 \, x_{3} - x_{4} &=0\\ -2 \, x_{1} + 6 \, x_{2} + 9 \, x_{3} + x_{4} &=0\\ -2 \, x_{1} + 11 \, x_{2} + 4 \, x_{3} - 9 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -2 & 7 & 8 & -1 \\ -3 & 10 & 12 & -1 \\ -2 & 6 & 9 & 1 \\ -2 & 11 & 4 & -9 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 1 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 31

V10 - Basis of Solution Space (ver. 31)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{2} + x_{3} - x_{4} &=0\\ -x_{1} - 4 \, x_{2} - 9 \, x_{3} + 3 \, x_{4} - 8 \, x_{5} &=0\\ x_{1} + 6 \, x_{2} + 12 \, x_{3} - 5 \, x_{4} + 10 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 0 & 1 & 1 & -1 & 0 \\ -1 & -4 & -9 & 3 & -8 \\ 1 & 6 & 12 & -5 & 10 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & 1 & -2 \\ 0 & 1 & 0 & -1 & -2 \\ 0 & 0 & 1 & 0 & 2 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ 2 \\ -2 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 32

V10 - Basis of Solution Space (ver. 32)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -5 \, x_{1} - 2 \, x_{2} - 5 \, x_{3} + 10 \, x_{4} &=0\\ -2 \, x_{1} - x_{2} - x_{3} + 5 \, x_{4} &=0\\ 3 \, x_{1} + 2 \, x_{2} - 9 \, x_{4} &=0\\ x_{1} - x_{2} + 8 \, x_{3} + 5 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -5 & -2 & -5 & 10 \\ -2 & -1 & -1 & 5 \\ 3 & 2 & 0 & -9 \\ 1 & -1 & 8 & 5 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 3 \\ 0 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 33

V10 - Basis of Solution Space (ver. 33)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + x_{2} + x_{3} + 2 \, x_{4} - 4 \, x_{5} &=0\\ x_{2} + 2 \, x_{3} + x_{4} - 2 \, x_{5} &=0\\ -x_{1} + 2 \, x_{2} + 5 \, x_{3} + x_{4} - 2 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 1 & 1 & 2 & -4 \\ 0 & 1 & 2 & 1 & -2 \\ -1 & 2 & 5 & 1 & -2 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & -1 & 1 & -2 \\ 0 & 1 & 2 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ -1 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ 2 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 34

V10 - Basis of Solution Space (ver. 34)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - 5 \, x_{2} - 2 \, x_{3} - 8 \, x_{4} &=0\\ -x_{1} + 6 \, x_{2} + 3 \, x_{3} + 8 \, x_{4} &=0\\ -x_{1} + 5 \, x_{2} + 3 \, x_{3} + 5 \, x_{4} &=0\\ -5 \, x_{2} - 6 \, x_{3} + 3 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & -5 & -2 & -8 \\ -1 & 6 & 3 & 8 \\ -1 & 5 & 3 & 5 \\ 0 & -5 & -6 & 3 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -3 \\ 3 \\ 1 \end{array}\right]\right\}.\]


Example 35

V10 - Basis of Solution Space (ver. 35)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - 2 \, x_{3} - 3 \, x_{4} &=0\\ x_{1} + x_{2} + 3 \, x_{3} + 9 \, x_{4} + 2 \, x_{5} &=0\\ 4 \, x_{1} - 7 \, x_{3} - 10 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 0 & -2 & -3 & 0 \\ 1 & 1 & 3 & 9 & 2 \\ 4 & 0 & -7 & -10 & 0 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 2 & 2 \\ 0 & 0 & 1 & 2 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ -2 \\ -2 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 0 \\ -2 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 36

V10 - Basis of Solution Space (ver. 36)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} 4 \, x_{2} - 8 \, x_{3} &=0\\ 2 \, x_{1} - x_{2} &=0\\ 2 \, x_{1} + 2 \, x_{2} - 6 \, x_{3} &=0\\ x_{1} - 5 \, x_{2} + 9 \, x_{3} &=0\\ x_{1} + x_{2} - 3 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 0 & 4 & -8 \\ 2 & -1 & 0 \\ 2 & 2 & -6 \\ 1 & -5 & 9 \\ 1 & 1 & -3 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ 2 \\ 1 \end{array}\right]\right\}.\]


Example 37

V10 - Basis of Solution Space (ver. 37)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} 0 &=0\\ x_{1} - 4 \, x_{2} - 6 \, x_{3} &=0\\ -x_{1} + 6 \, x_{2} + 8 \, x_{3} &=0\\ -x_{1} + 2 \, x_{2} + 4 \, x_{3} &=0\\ x_{1} - 5 \, x_{2} - 7 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 0 & 0 & 0 \\ 1 & -4 & -6 \\ -1 & 6 & 8 \\ -1 & 2 & 4 \\ 1 & -5 & -7 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ -1 \\ 1 \end{array}\right]\right\}.\]


Example 38

V10 - Basis of Solution Space (ver. 38)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 4 \, x_{2} + 7 \, x_{3} - 5 \, x_{4} + 2 \, x_{5} &=0\\ x_{1} + 5 \, x_{2} + 9 \, x_{3} - 6 \, x_{4} + 3 \, x_{5} &=0\\ 0 &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 4 & 7 & -5 & 2 \\ 1 & 5 & 9 & -6 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & -1 & -1 & -2 \\ 0 & 1 & 2 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} 1 \\ 1 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ -1 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 39

V10 - Basis of Solution Space (ver. 39)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -5 \, x_{2} &=0\\ -2 \, x_{1} - 9 \, x_{2} + 4 \, x_{3} &=0\\ -2 \, x_{1} - 12 \, x_{2} + 4 \, x_{3} &=0\\ -x_{1} - 10 \, x_{2} + 2 \, x_{3} &=0\\ -x_{1} - 5 \, x_{2} + 2 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 0 & -5 & 0 \\ -2 & -9 & 4 \\ -2 & -12 & 4 \\ -1 & -10 & 2 \\ -1 & -5 & 2 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 40

V10 - Basis of Solution Space (ver. 40)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} - 4 \, x_{2} - 5 \, x_{3} + 6 \, x_{4} &=0\\ 2 \, x_{1} + 8 \, x_{2} + 7 \, x_{3} - 9 \, x_{4} &=0\\ 2 \, x_{1} + 8 \, x_{2} + 8 \, x_{3} - 10 \, x_{4} &=0\\ x_{1} + 4 \, x_{2} + 4 \, x_{3} - 5 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -1 & -4 & -5 & 6 \\ 2 & 8 & 7 & -9 \\ 2 & 8 & 8 & -10 \\ 1 & 4 & 4 & -5 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 4 & 0 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -4 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} 1 \\ 0 \\ 1 \\ 1 \end{array}\right]\right\}.\]


Example 41

V10 - Basis of Solution Space (ver. 41)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} 2 \, x_{1} - x_{2} - x_{3} - 10 \, x_{4} + 3 \, x_{5} &=0\\ x_{1} - 3 \, x_{4} + x_{5} &=0\\ 2 \, x_{1} + x_{3} - 4 \, x_{4} + x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 2 & -1 & -1 & -10 & 3 \\ 1 & 0 & 0 & -3 & 1 \\ 2 & 0 & 1 & -4 & 1 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & -3 & 1 \\ 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & 2 & -1 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 3 \\ -2 \\ -2 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 42

V10 - Basis of Solution Space (ver. 42)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{2} - 2 \, x_{3} - 4 \, x_{4} &=0\\ -x_{1} + x_{2} + 2 \, x_{3} + x_{4} &=0\\ -x_{2} + 2 \, x_{3} + 5 \, x_{4} &=0\\ x_{2} - 2 \, x_{3} - 5 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 0 & 1 & -2 & -4 \\ -1 & 1 & 2 & 1 \\ 0 & -1 & 2 & 5 \\ 0 & 1 & -2 & -5 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & -4 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 4 \\ 2 \\ 1 \\ 0 \end{array}\right]\right\}.\]


Example 43

V10 - Basis of Solution Space (ver. 43)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 4 \, x_{2} - 3 \, x_{3} + 7 \, x_{4} - 7 \, x_{5} &=0\\ x_{2} - 2 \, x_{3} &=0\\ -x_{1} - 8 \, x_{2} + 12 \, x_{3} - 6 \, x_{4} + 6 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 4 & -3 & 7 & -7 \\ 0 & 1 & -2 & 0 & 0 \\ -1 & -8 & 12 & -6 & 6 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & 2 & -2 \\ 0 & 1 & 0 & 2 & -2 \\ 0 & 0 & 1 & 1 & -1 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -2 \\ -2 \\ -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ 2 \\ 1 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 44

V10 - Basis of Solution Space (ver. 44)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} + 2 \, x_{3} + 2 \, x_{5} &=0\\ x_{1} + x_{2} + x_{3} - 3 \, x_{4} + 2 \, x_{5} &=0\\ 5 \, x_{1} + 3 \, x_{2} + 7 \, x_{3} - 9 \, x_{4} + 10 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} 1 & 0 & 2 & 0 & 2 \\ 1 & 1 & 1 & -3 & 2 \\ 5 & 3 & 7 & -9 & 10 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 2 & 0 & 2 \\ 0 & 1 & -1 & -3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -2 \\ 1 \\ 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{r} 0 \\ 3 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -2 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 45

V10 - Basis of Solution Space (ver. 45)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -3 \, x_{1} + 7 \, x_{2} - x_{3} &=0\\ 2 \, x_{1} - 5 \, x_{2} + x_{3} &=0\\ -2 \, x_{1} + 9 \, x_{2} - 5 \, x_{3} &=0\\ 3 \, x_{1} - 9 \, x_{2} + 3 \, x_{3} &=0\\ 2 \, x_{1} - 8 \, x_{2} + 4 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} -3 & 7 & -1 \\ 2 & -5 & 1 \\ -2 & 9 & -5 \\ 3 & -9 & 3 \\ 2 & -8 & 4 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ 1 \\ 1 \end{array}\right]\right\}.\]


Example 46

V10 - Basis of Solution Space (ver. 46)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} + x_{2} - 4 \, x_{3} + 5 \, x_{4} &=0\\ -x_{1} - 4 \, x_{3} + 7 \, x_{4} &=0\\ 2 \, x_{2} + x_{3} - 6 \, x_{4} &=0\\ -x_{1} - 2 \, x_{2} + 3 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} -1 & 1 & -4 & 5 \\ -1 & 0 & -4 & 7 \\ 0 & 2 & 1 & -6 \\ -1 & -2 & 0 & 3 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} -1 \\ 2 \\ 2 \\ 1 \end{array}\right]\right\}.\]


Example 47

V10 - Basis of Solution Space (ver. 47)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - x_{2} &=0\\ 2 \, x_{1} - 7 \, x_{2} + 10 \, x_{3} &=0\\ x_{1} - 2 \, x_{3} &=0\\ -x_{2} + 2 \, x_{3} &=0\\ x_{1} - 6 \, x_{2} + 10 \, x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrr} 1 & -1 & 0 \\ 2 & -7 & 10 \\ 1 & 0 & -2 \\ 0 & -1 & 2 \\ 1 & -6 & 10 \end{array}\right]\sim\left[\begin{array}{rrr} 1 & 0 & -2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 2 \\ 2 \\ 1 \end{array}\right]\right\}.\]


Example 48

V10 - Basis of Solution Space (ver. 48)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} -x_{1} - 4 \, x_{2} - 2 \, x_{3} - 8 \, x_{4} + 10 \, x_{5} &=0\\ x_{2} - x_{3} + 5 \, x_{4} - 2 \, x_{5} &=0\\ x_{1} + 4 \, x_{2} + x_{3} + 10 \, x_{4} - 10 \, x_{5} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrrr} -1 & -4 & -2 & -8 & 10 \\ 0 & 1 & -1 & 5 & -2 \\ 1 & 4 & 1 & 10 & -10 \end{array}\right]\sim\left[\begin{array}{rrrrr} 1 & 0 & 0 & 0 & -2 \\ 0 & 1 & 0 & 3 & -2 \\ 0 & 0 & 1 & -2 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 0 \\ -3 \\ 2 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 2 \\ 2 \\ 0 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 49

V10 - Basis of Solution Space (ver. 49)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} x_{1} - x_{3} - 3 \, x_{4} &=0\\ -x_{1} + x_{2} + 2 \, x_{3} + 5 \, x_{4} &=0\\ -4 \, x_{1} - 3 \, x_{2} + x_{3} + 6 \, x_{4} &=0\\ -4 \, x_{1} - 5 \, x_{2} - x_{3} + 2 \, x_{4} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 1 & 0 & -1 & -3 \\ -1 & 1 & 2 & 5 \\ -4 & -3 & 1 & 6 \\ -4 & -5 & -1 & 2 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & -1 & -3 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 3 \\ -2 \\ 0 \\ 1 \end{array}\right]\right\}.\]


Example 50

V10 - Basis of Solution Space (ver. 50)

Explain how to find a basis for the solution space of the homogeneous system \begin{align*} 4 \, x_{1} - 8 \, x_{2} + 4 \, x_{3} &=0\\ 2 \, x_{1} - 3 \, x_{2} + x_{3} + x_{4} &=0\\ -x_{1} + 7 \, x_{2} - 6 \, x_{3} + 5 \, x_{4} &=0\\ -x_{1} + 2 \, x_{2} - x_{3} &=0\\ \end{align*}

Answer.

\[\left[\begin{array}{rrrr} 4 & -8 & 4 & 0 \\ 2 & -3 & 1 & 1 \\ -1 & 7 & -6 & 5 \\ -1 & 2 & -1 & 0 \end{array}\right]\sim\left[\begin{array}{rrrr} 1 & 0 & -1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\]

A basis is

\[\left\{\left[\begin{array}{r} 1 \\ 1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -2 \\ -1 \\ 0 \\ 1 \end{array}\right]\right\}.\]