## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} 1 & 1 & -1 \\ -3 & 2 & -1 \\ 3 & -3 & 2 \end{bmatrix}$
The general form of a matrix of order $3 \times 3$ is: $det \ A=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h& i \end{bmatrix}=a(ei-fh) -b(di-fg)+c(dh-eg)$ Now, $det \ A=\begin{bmatrix} 1 & 1 &1 \\ 3 & 5 & 4 \\ 3 &6 & 5 \end{bmatrix}=1 (25-24)- 1 (15-12)+1 (18-15)=1$ and $A^{-1}=\dfrac{1}{1} \begin{bmatrix} 1 & 1 & -1 \\ -3 & 2 & -1 \\ 3 & -3 & 2 \end{bmatrix}=\begin{bmatrix} 1 & 1 & -1 \\ -3 & 2 & -1 \\ 3 & -3 & 2 \end{bmatrix}$ Our answer is: $A^{-1}=\begin{bmatrix} 1 & 1 & -1 \\ -3 & 2 & -1 \\ 3 & -3 & 2 \end{bmatrix}$