## Algebra and Trigonometry 10th Edition

The general form of a matrix of order $3 \times 3$ is: $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h& i \end{bmatrix}=a(ei-fh) -b(di-fg)+c(dh-eg)$ Now, $det \ A=\begin{bmatrix} 1 & 0 & 0 \\ 3 & 0 & 0 \\ 2 & 5 & 5 \end{bmatrix} \ne 1$ Multiply row $2$ by $\dfrac{1}{3}$ and then subtract row 2 from row 1. Thus, $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 5 & 5 \end{bmatrix}$ Since the matrix is neither invertible nor singular, the inverse of the matrix does not exist.