Find A⁻¹, if A = (1 2 0; -2 -1 -2; 0 -1 1). Using A⁻¹, solve the system of linear equations: x - 2y…

$|A| = 1(-1 - 2) - 2(-2 - 0) + 0 = -3 + 4 = 1 \neq 0$, so $A^{-1}$ exists. Cofactors: $C_{11} = -3$, $C_{12} = -2$, $C_{13} = 2$, $C_{21} = -2$, $C_{22} = 1$, $C_{23} = 1$, $C_{31} = -4$, $C_{32} = 2$, $C_{33} = 3$ $\text{adj } A = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix}$ $A^{-1} = \frac{1}{|A|}\text{adj } A = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix}$ The system is $A^T X = B$, where $A^T = \begin{pmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, $B = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix}$. $X = (A^T)^{-1}B = (A^{-1})^T B = \begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix}\begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -3 \end{pmatrix}$ Hence $x = 0$, $y = -5$, $z = -3$.