The solution of the system of equations 2x + y - z = 3 x - y + 2z = 5…

The correct answer is Option (4) → \(x = 2,\ y = 1,\ z = 2\)

Write the system as AX = B with

    \( \displaystyle A = \begin{bmatrix}2 & 1 & -1\\ 1 & -1 & 2\\ 1 & 2 & 1\end{bmatrix},\quad X = \begin{bmatrix}x\\ y\\ z\end{bmatrix},\quad B = \begin{bmatrix}3\\ 5\\ 6\end{bmatrix} \)

Cofactors of the first row:

    \(C_{11} = (-1)(1) - (2)(2) = -5\)
    \(C_{12} = -\big[(1)(1) - (2)(1)\big] = 1\)
    \(C_{13} = (1)(2) - (-1)(1) = 3\)

So

    \(|A| = 2(-5) + 1(1) + (-1)(3) = -12\)

Since \(|A| \ne 0\), the system has a unique solution \(X = A^{-1}B\). The other cofactors:

    \(C_{21} = -\big[(1)(1) - (-1)(2)\big] = -3\)
    \(C_{22} = (2)(1) - (-1)(1) = 3\)
    \(C_{23} = -\big[(2)(2) - (1)(1)\big] = -3\)

    \(C_{31} = (1)(2) - (-1)(-1) = 1\)
    \(C_{32} = -\big[(2)(2) - (-1)(1)\big] = -5\)
    \(C_{33} = (2)(-1) - (1)(1) = -3\)

The adjoint is the transpose of the cofactor matrix:

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

Then

    \( \displaystyle X = \frac{1}{-12}\begin{bmatrix}-5 & -3 & 1\\ 1 & 3 & -5\\ 3 & -3 & -3\end{bmatrix}\begin{bmatrix}3\\ 5\\ 6\end{bmatrix} \)

    \( \displaystyle X = \frac{1}{-12}\begin{bmatrix}-15 - 15 + 6\\ 3 + 15 - 30\\ 9 - 15 - 18\end{bmatrix} \)

    \( \displaystyle X = \frac{1}{-12}\begin{bmatrix}-24\\ -12\\ -24\end{bmatrix} = \begin{bmatrix}2\\ 1\\ 2\end{bmatrix} \)

Check in the first equation: \(2(2) + 1 - 2 = 3\).