Question

Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimination. -x+y-z=-20,2x-y+z=29, 3x+2y+z=29

Answer 1

In matrix form, the system is given by

`\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}`

I'll use G-J elimination. Consider the augmented matrix

`\left[ \begin{array}c -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]`

• Multiply through row 1 by -1.

`\left[ \begin{array}c 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]`

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

`\left[ \begin{array}c 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]`

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

`\left[ \begin{array}c 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]`

• Multiply row 3 by 1/3.

`\left[ \begin{array}c 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]`

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

`\left[ \begin{array}ccc 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]`

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

`\left[ \begin{array}c 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]`

Then the solution to the system is

`\boxed{x=9, y=-3, z=8}`

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

`\left[ \begin{array}c 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]`

The third row tells us that
`z=8`. Then in the second row,

`y-z = -11 \implies y=-11 + 8 = -3`

and in the first row,

`x-y+z=20 \implies x=20 + (-3) - 8 = 9`