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