[tex]A^{-1}\text{ =}\begin{bmatrix}{2} & {-5} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}\text{ (option B)}[/tex]
Explanation:[tex]\begin{bmatrix}{3} & {5} & {} \\ {1} & {2} & {} \\ {} & {} & {}\end{bmatrix}[/tex]
To find the inverse of the matrix, first let's find the determinant:
[tex]\begin{gathered} |A|\text{ = 3(2) - 5(1)} \\ |A|\text{ = 6 - 5} \\ |A|\text{ = 1} \end{gathered}[/tex]
Then, we'll find the Adjunct of the matrix:
[tex]\begin{gathered} \begin{bmatrix}{3} & {5} & {} \\ {1} & {2} & {} \\ {} & {} & {}\end{bmatrix}\text{ : interchange }3\text{ and 2. negate 1 and 5} \\ \text{Adjunct = }\begin{bmatrix}{2} & {-5} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}[/tex][tex]\begin{gathered} In\text{verse of the matrix = }\frac{1}{|A|}\times\text{ adjunct} \\ A^{-1}\text{ = }\frac{1}{1}(\begin{bmatrix}{2} & {-5} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}) \\ A^{-1}\text{ =}\begin{bmatrix}{2} & {-5} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}\text{ (option B)} \\ \end{gathered}[/tex]
