# Solve the system of equations using matrices.Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. begin{cases}x+y-z=-22x-y+z=5-x+2y+2z=1end{cases}

Solve the system of equations using matrices.Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
$$\begin{cases}x+y-z=-2\\2x-y+z=5\\-x+2y+2z=1\end{cases}$$

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Step 1
Given:
The system of linear equations are,
$$x+y-z=-2$$
$$2x-y+z=5$$
$$-x+2y+2z=1$$
For applying Gauss-Jordan elimination method, the above system of equations can be represented in matrix form as,
$$\begin{bmatrix}1 & 1&-1&-2 \\2 & -1&1&5\\-1&2&2&1 \end{bmatrix}$$
Step 2
The above matrix can be converted into row echelon form as,
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & -3&3&9\\-1&2&2&1 \end{bmatrix} \left(\text{By applying: } R_2 \rightarrow R_2-2R_1\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & -3&3&9\\0&3&1&-1 \end{bmatrix}\left(\text{By applying: } R_3 \rightarrow R_3+R_1\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&-1&-3\\0&3&1&-1 \end{bmatrix} \left(\text{By applying: } R_2 \rightarrow \frac{-1}{3}R_2\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&-1&-3\\0&0&4&8 \end{bmatrix} \left(\text{By applying: } R_3 \rightarrow R_3-3R_2\right)$$
Step 3
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&-1&-3\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_3 \rightarrow \frac{1}{4}R_3\right)$$
$$\begin{bmatrix}1 & 1&-1&-2 \\0 & 1&0&-1\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_2 \rightarrow R_2+R_3\right)$$
$$\begin{bmatrix}1 & 1&0&0 \\0 & 1&0&-1\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_1 \rightarrow R_1+R_3\right)$$
Step 4
$$\begin{bmatrix}1 & 0&0&1 \\0 & 1&0&-1\\0&0&1&2 \end{bmatrix} \ \ \left(\text{By applying: } R_1 \rightarrow R_1-R_2\right)$$
Hence, the solutions of the given system of equations are x=1 , y=-1 and z=2