Solution
Step 1:
Write the matrix:
[tex]\begin{gathered} X\text{ = }\begin{bmatrix}{6} & {a} \\ {-2} & {-5}\end{bmatrix} \\ I\text{ = }\begin{bmatrix}{1} & {0} \\ {0} & {1}\end{bmatrix} \end{gathered}[/tex]Step 2:
Find the inverse of X.
[tex]X^{-1}\text{ = }\frac{Adjoint}{Determinant}[/tex]Step 3:
Find the determinant
[tex]\begin{gathered} Determinant\text{ of X = }\begin{bmatrix}{6} & {a} \\ {-2} & {-5}\end{bmatrix} \\ Co-factor\text{ of X = }\begin{bmatrix}{-5} & {2} \\ {-a} & {6}\end{bmatrix} \\ Adjoint\text{ is the transpose of the co-factor of X} \\ Adj\text{oint = }\begin{bmatrix}{-5} & {-a} \\ {2} & {6}\end{bmatrix} \\ Deterninant\text{ = 2a - 30} \end{gathered}[/tex]Step 4:
[tex]\begin{gathered} X\text{ - }X^{-1}\text{ = I} \\ \\ \begin{bmatrix}{6} & {a} \\ {-2} & {-5}\end{bmatrix}\text{ - }\frac{\begin{bmatrix}{-5} & {-a} \\ {2} & {6}\end{bmatrix}}{2a\text{ - 30}}\text{ = }\begin{bmatrix}{1} & {0} \\ {0} & {1}\end{bmatrix} \\ \begin{bmatrix}{6} & {a} \\ {-2} & {-5}\end{bmatrix}\text{ - }\begin{bmatrix}{\frac{-5}{2a-30}} & {\frac{-a}{2a-30}} \\ {\frac{2}{2a-30}} & {\frac{6}{2a-30}}\end{bmatrix}\text{= }\begin{bmatrix}{1} & {0} \\ {0} & {1}\end{bmatrix} \\ \begin{bmatrix}{6\text{ + }\frac{5}{2a-30}} & {a\text{ + }\frac{a}{2a\text{ - 30}}} \\ {-2\text{ -}\frac{2}{2a\text{ - 30}}} & {-5-\text{ }\frac{6}{2a\text{ - 30}}}\end{bmatrix}\text{ = }\begin{bmatrix}{1} & {0} \\ {0} & {1}\end{bmatrix} \end{gathered}[/tex]Step 5;
Write equations
[tex]\begin{gathered} \text{ 6 + }\frac{5}{2a\text{ - 30 }}\text{ = 1} \\ \frac{6(2a\text{ - 30\rparen + 5}}{2a\text{ - 30}}\text{ = 1} \\ \frac{6(2a\text{ -30\rparen + 5}}{2a\text{ - 30}}\text{ = 1} \\ 12a\text{ - 180 + 5 = 2a - 30} \\ 12a\text{ - 2a = 175 - 30} \\ 10a\text{ = 145} \\ \text{a = }\frac{145}{10} \\ \text{a = }\frac{29}{2} \end{gathered}[/tex]Final answer
[tex]\begin{gathered} i)\text{ }X^{-1}\text{ = }\frac{1}{2a\text{ - 30}}\begin{bmatrix}{-5} & {-a} \\ {2} & {6}\end{bmatrix} \\ ii)\text{ }a\text{ = }\frac{29}{2}\text{ or 14.5} \end{gathered}[/tex]
