Question

#24, i am getting
`\left[\begin{array}{cccc}1&0&0&(18)/(7) \\0&1&0&(8)/(7) \\0&0&0&(2)/(7) \end{array}\right]` , im wondering if someone can check it and show their work so i may compare.

Answer 1

It looks like you're talking about row-reducing an augmented matrix to solve the system of equations. Your answer is almost correct. The last row should read 0, 0, 1, 2/7.

The given system translates to

`\left[ \begin{array}ccc 2 & -3 & 1 & 2 \\ 1 & -1 & 2 & 2 \\ 1 & 2 & -3 & 4 \end{array} \right]`

Eliminate x from the last two rows by combining -2 (row 2) and row 1, and -2 (row 3) and row 1; that is,

(2x - 3y + z) - 2 (x - y + 2z) = 2 - 2 (2)

2x - 3y + z - 2x + 2y - 4z = 2 - 4

-y - 3z = -2

and

(2x - 3y + z) - 2 (x + 2y - 3z) = 2 - 2 (4)

2x - 3y + z - 2x - 4y + 6z = 2 - 8

-7y + 7z = -6

In augmented matrix form, this step yields

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

I'll omit these details in the remaining steps.

Eliminate y from the last row by combining -7 (row 2) and row 3 :

`\left[ \begin{array}ccc 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 28 & 8 \end{array} \right]`

Multiply the last row by 1/28 :

`\left[ \begin{array}ccc 2 & -3 & 1 & 2 \\ 0 & -1 & -3 & -2 \\ 0 & 0 & 1 & 2/7 \end{array} \right]`

Eliminate z from the second row by combining 3 (row 3) and row 2 :

`\left[ \begin{array}c 2 & -3 & 1 & 2 \\ 0 & -1 & 0 & -8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]`

Multiply the second row by -1 :

`\left[ \begin{array}ccc 2 & -3 & 1 & 2 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]`

Eliminate y and z from the first row by combining 3 (row 2), -1 (row 3), and row 1 :

`\left[ \begin{array}c 2 & 0 & 0 & 36/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]`

Multiply the first row by 1/2 :

`\left[ \begin{array}c 1 & 0 & 0 & 18/7 \\ 0 & 1 & 0 & 8/7 \\ 0 & 0 & 1 & 2/7 \end{array} \right]`