The solution of the system of equations x + 2y + z = 3 2x - y + z = 8…

The correct answer is Option (3) → \(x = 2,\ y = -1,\ z = 3\)

Write the system as AX = B with

    \( \displaystyle A = \begin{bmatrix}1 & 2 & 1\\ 2 & -1 & 1\\ 1 & 1 & -2\end{bmatrix},\quad B = \begin{bmatrix}3\\ 8\\ -5\end{bmatrix} \)

The cofactors of A are

    \( C_{11} = 2 - 1 = 1,\ C_{12} = -(-4 - 1) = 5,\ C_{13} = 2 + 1 = 3 \)

    \( C_{21} = -(-4 - 1) = 5,\ C_{22} = -2 - 1 = -3,\ C_{23} = -(1 - 2) = 1 \)

    \( C_{31} = 2 + 1 = 3,\ C_{32} = -(1 - 2) = 1,\ C_{33} = -1 - 4 = -5 \)

Expanding along the first row:

    \( |A| = 1(1) + 2(5) + 1(3) = 14 \ne 0 \)

so the system has a unique solution \(X = A^{-1}B\). The adjoint is the transpose of the cofactor matrix:

    \( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}1 & 5 & 3\\ 5 & -3 & 1\\ 3 & 1 & -5\end{bmatrix} \)

Then

    \( \displaystyle X = \frac{1}{14}\begin{bmatrix}1 & 5 & 3\\ 5 & -3 & 1\\ 3 & 1 & -5\end{bmatrix}\begin{bmatrix}3\\ 8\\ -5\end{bmatrix} \)

    \( \displaystyle = \frac{1}{14}\begin{bmatrix}3 + 40 - 15\\ 15 - 24 - 5\\ 9 + 8 + 25\end{bmatrix} \)

    \( \displaystyle = \frac{1}{14}\begin{bmatrix}28\\ -14\\ 42\end{bmatrix} = \begin{bmatrix}2\\ -1\\ 3\end{bmatrix} \)

So x = 2, y = −1, z = 3.

Shortcut: Substituting each option into all three equations also works. Only Option (3) satisfies all three; Options (1), (2) and (4) satisfy the first equation but fail the second.