Question

Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.

Answer 1

In augmented matrix form, the system is equivalent to

$\begin{bmatrix}1&1&1&-5\\1&-1&3&-1\\4&1&1&-2\end{bmatrix}$

Subtract row 1 from row 2, and 4(row 1) from row 3:

$\begin{bmatrix}1&1&1&-5\\0&-2&2&4\\0&-3&-3&18\end{bmatrix}$

Divide through row 2 by -2, and through row 3 by -3:

$\begin{bmatrix}1&1&1&-5\\0&1&-1&-2\\0&1&1&-6\end{bmatrix}$

Subtract row 2 from row 3:

$\begin{bmatrix}1&1&1&-5\\0&1&-1&-2\\0&0&2&-4\end{bmatrix}$

Divide through row 3 by 2:

$\begin{bmatrix}1&1&1&-5\\0&1&-1&-2\\0&0&1&-2\end{bmatrix}$

The last row tells you $\boxed{z=-2}$. Then

$y-z=-2\implies y+2=-2\implies\boxed{y=-4}$

and

$x+y+z=-5\implies x-4-2=-5\implies\boxed{x=1}$