Q.Solve the system $x+y+z=6$, $\ 2x+3y-z=5$, $\ 3x-y+z=4$ by the reduction method.

Form the augmented matrix $[\,A\mid B\,]$, reduce it to upper-triangular form by row operations, then back-substitute.

Augmented matrix. $[\,A\mid B\,]=\left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 3 & -1 & 5 \\ 3 & -1 & 1 & 4 \end{array}\right).$

$R_2\to R_2-2R_1$ and $R_3\to R_3-3R_1$: $R_2:(0,\ 1,\ -3\mid 5-12=-7),\qquad R_3:(0,\ -4,\ -2\mid 4-18=-14).$ $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -3 & -7 \\ 0 & -4 & -2 & -14 \end{array}\right).$

$R_3\to R_3+4R_2$: $R_3:(0,\ 0,\ -2-12=-14\mid -14+4(-7)=-42).$ $\left(\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -3 & -7 \\ 0 & 0 & -14 & -42 \end{array}\right).$ …

The system could also be solved by matrix inversion (X = A^{-1}B) or by Cramer's Rule; the reduction method is shown …

The usual errors are applying a row operation to the coefficient side but forgetting the constant on the right, and slips in the arithmetic of R_i -> R_i + kR_j. Carry the constant column through every o …