Q.Solve the system $x+2y+3z=14$, $\ 2x-y+z=3$, $\ 3x+y-z=2$ by the reduction method.

Reduce the augmented matrix to upper-triangular form, then back-substitute.

Augmented matrix. $\left(\begin{array}{ccc|c} 1 & 2 & 3 & 14 \\ 2 & -1 & 1 & 3 \\ 3 & 1 & -1 & 2 \end{array}\right).$

$R_2\to R_2-2R_1$ and $R_3\to R_3-3R_1$: $R_2:(0,\ -5,\ -5\mid 3-28=-25),\qquad R_3:(0,\ -5,\ -10\mid 2-42=-40).$ $\left(\begin{array}{ccc|c} 1 & 2 & 3 & 14 \\ 0 & -5 & -5 & -25 \\ 0 & -5 & -10 & -40 \end{array}\right).$

$R_3\to R_3-R_2$: $R_3:(0,\ 0,\ -10-(-5)=-5\mid -40-(-25)=-15).$ $\left(\begin{array}{ccc|c} 1 & 2 & 3 & 14 \\ 0 & -5 & -5 & -25 \\ 0 & 0 & -5 & -15 \end{array}\right).$ …

The same solution follows by matrix inversion (X = A^{-1}B) or Cramer's Rule; reduction is chose …

The most frequent errors are dropping the constant column during a row operation and sign slips in R_3 -> R_3 - R_2. Keep the constants attached and, at the …