Question

Solve by using matrices. 2x – y +2 + w = -3 x + 2y – 3z + w = 12 3x - y - + 2w = 3 -2x + 3y + 2 – 3w = -3​

Answer 1

Some symbols and numbers are missing. I assume the system is supposed to read

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

x + 2y - 3z + w = 12

3x - y - z + 2w = 3

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

In matrix form, this is

`\begin{bmatrix}2&-1&2&1\\1&2&-3&1\\3&-1&-1&2\\-2&3&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}-3\\12\\3\-3\end{bmatrix}`

which we can strip down to the augmented matrix,

`\left[\begin{array}c2&-1&2&1&-3\\1&2&-3&1&12\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right]`

Now for the row operations:

• swap rows 1 and 2

`\left[\begin{array}cccc1&2&-3&1&12\\2&-1&2&1&-3\\3&-1&-1&2&3\\-2&3&2&-3&-3\end{array}\right]`

• add -2 (row 1) to row 2, -3 (row 1) to row 3, and 2 (row 1) to row 4

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&-7&8&-1&-33\\0&7&-4&-1&21\end{array}\right]`

• add 7 (row 2) to -5 (row 3), and row 3 to row 4

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&16&-2&-24\\0&0&4&-2&-12\end{array}\right]`

• multiply through rows 3 and 4 by 1/2

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&8&-1&-12\\0&0&2&-1&-6\end{array}\right]`

• add -4 (row 4) to row 3

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&0&3&12\\0&0&2&-1&-6\end{array}\right]`

• swap rows 3 and 4

`\left[\begin{array}cccc1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&3&12\end{array}\right]`

• multiply through row 4 by 1/3

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&-1&-6\\0&0&0&1&4\end{array}\right]`

• add row 4 to row 3

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&2&0&-2\\0&0&0&1&4\end{array}\right]`

• multiply through row 3 by 1/2

`\left[\begin{array}c1&2&-3&1&12\\0&-5&8&-1&-27\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]`

• add -8 (row 3) and row 4 to row 2

`\left[\begin{array}cccc1&2&-3&1&12\\0&-5&0&0&-15\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]`

• multiply through row 2 by -1/5

`\left[\begin{array}cccc1&2&-3&1&12\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]`

• add -2 (row 2) and 3 (row 3) and -1 (row 4) to row 1

`\left[\begin{array}c1&0&0&0&-1\\0&1&0&0&3\\0&0&1&0&-1\\0&0&0&1&4\end{array}\right]`

Then the solution to the system is (x, y, z, w) = (-1, 3, -1, 4).