Question

Help with matrices!!!

Answer 1

`\begin{cases}x+2y+z=5\\2x-y+2z=15\\3x+y-z=8\end{cases}`

First, write the system in matrix form:

`\begin{bmatrix}1&2&1\\2&-1&2\\3&1&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\15\\8\end{bmatrix}`

Now take the augmented matrix,

`\left[\begin{array}c1&2&1&5\\2&-1&2&15\\3&1&-1&8\end{array}\right]`

Add -2(row 1) to row 2, and add -3(row 1) to row 3:

`\left[\begin{array}ccc1&2&1&5\\0&-5&0&5\\0&-5&-4&-7\end{array}\right]`

Add -1(row 2) to row 3:

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

Multiply through row 2 by -1/5, and multiply through row 3 by -1/4:

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

We can already solve for
`y` and
`z` from here, but we'll hold off on that because the matrix still isn't in RREF, since the leftmost element in row 2 (1) is not the only non-zero element in its column. We fix this by adding -2(row 2) to row 1:

`\left[\begin{array}c1&0&1&7\\0&1&0&-1\\0&0&1&3\end{array}\right]`

and adding -1(row 3) to row 1:

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

This tells us the solution to this system is
`(x,y,z)=\boxed{(4,-1,3)}`.