Question

13. Solve the system using the elimination method. x + 3y - z = 2x + y - z = 0 3x + 2y - 3z =-1

Answer 1

We have the following:

$\begin{gathered} x+3y-z=2 \\ x+y-z=0 \\ 3x+2y-3z=-1 \end{gathered}$

solving for elimination:

$\begin{bmatrix}{1} & {3} & -1 \\ {1} & {1} & {1} \\ {3} & {2} & {-3}\end{bmatrix}=\begin{aligned}2 \\ 0 \\ -1\end{aligned}$

R1 <-> R3

$\begin{bmatrix}{3} & {2} & -3 \\ {1} & {1} & {1} \\ {1} & {3} & {-1}\end{bmatrix}=\begin{aligned}-1 \\ 0 \\ 2\end{aligned}$

R2 - 1/3*R1

$\begin{bmatrix}{3} & {2} & -3 \\ {0} & {\frac{1}{3}} & {0} \\ {1} & {3} & {-1}\end{bmatrix}=\begin{aligned}-1 \\ \frac{1}{3} \\ 2\end{aligned}$

R3 - 1/3 * R1

$\begin{bmatrix}{3} & {2} & -3 \\ {0} & {\frac{1}{3}} & {0} \\ {0} & {\frac{7}{3}} & {0}\end{bmatrix}=\begin{aligned}-1 \\ \frac{1}{3} \\ \frac{7}{3}\end{aligned}$

R2 <-> R3

$\begin{bmatrix}{3} & {2} & -3 \\ {0} & {\frac{7}{3}} & {0} \\ {0} & {\frac{1}{3}} & {0}\end{bmatrix}=\begin{aligned}-1 \\ \frac{7}{3} \\ \frac{1}{3}\end{aligned}$

R3 - 1/7*R2

$\begin{bmatrix}{3} & {2} & -3 \\ {0} & {\frac{7}{3}} & {0} \\ {0} & {0} & {0}\end{bmatrix}=\begin{aligned}-1 \\ \frac{7}{3} \\ 0\end{aligned}$

Zero row in reduced matrix indicates infinite solutions