आव्यूह का प्रयोग करके मान ज्ञात कीजिये? x+y+z=6, x−y+z=2, 2x+y−z=1

$$|A| = 1 \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix}$$ $$|A| = 1((-1)(-1) - (1)(1)) - 1((1)(-1) - (1)(2)) + 1((1)(1) - (-1)(2))$$ $$|A| = 1(1 - 1) - 1(-1 - 2) + 1(1 + 2)$$ $$|A| = 1(0) - 1(-3) + 1(3)$$ $$|A| = 0 + 3 + 3 = 6$$

चूंकि $|A| \neq 0$, इसलिए $A^{-1}$ का अस्तित्व है।

पहले सहखंड आव्यूह (cofactor matrix) $C$ ज्ञात करें:
$C_{11} = 0$
$C_{12} = 3$
$C_{13} = 3$
$C_{21} = 2$
$C_{22} = -3$
$C_{23} = 1$
$C_{31} = 2$
$C_{32} = 0$
$C_{33} = -2$

$$C = \begin{bmatrix} 0 & 3 & 3 \\ 2 & -3 & 1 \\ 2 & 0 & -2 \end{bmatrix}$$

सहखंडज आव्यूह $adj(A) = C^T$ है:

$$adj(A) = \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix}$$

$$A^{-1} = \frac{1}{|A|} adj(A)$$ $$A^{-1} = \frac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix}$$

$$X = \frac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix} \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix}$$ $$X = \frac{1}{6} \begin{bmatrix} (0)(6) + (2)(2) + (2)(1) \\ (3)(6) + (-3)(2) + (0)(1) \\ (3)(6) + (1)(2) + (-2)(1) \end{bmatrix}$$ $$X = \frac{1}{6} \begin{bmatrix} 0 + 4 + 2 \\ 18 - 6 + 0 \\ 18 + 2 - 2 \end{bmatrix}$$ $$X = \frac{1}{6} \begin{bmatrix} 6 \\ 12 \\ 18 \end{bmatrix}$$ $$X = \begin{bmatrix} 6/6 \\ 12/6 \\ 18/6 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$

इसलिए, $x=1, y=2, z=3$.