Q.Given $A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$, compute $A^{-1}$ and show that $2A^{-1} = 9I - A$.

For a square matrix $A$ of order $m$, another square matrix $B$ of the same order is called the inverse of $A$ if $AB = BA = I$, the identity matrix of order $m$; $B$ is written $A^{-1}$, and by the same relation $A$ is equally the inverse of $B$, so $A=B^{-1}$. A matrix is called invertible exactly when such a $B$ exists, which happens if and only if $A$ is non-singular, $|A|\neq0$. The inverse, when it exists, is always unique: if $B$ and $C$ were both inverses of $A$, then $B = BI = B(AC) = (BA)C = IC = C$, so $B=C$. Two standard routes compute $A^{-1}$ — the elementary-transformation method and the adjoint-formula method — and both must agree, since the inverse is unique. Beyond direct computation, the defining relation $AA^{-1}=A^{-1}A=I$ is itself a powerful algebraic tool: a matrix equation such as $AX=B$ (with $A$ square and non-singular) is solved by pre-multiplying both sides by $A^{-1}$, giving $X=A^{-1}B$; an equation $XA=B$ is instead solved by post-multiplying by $A^{-1}$, giving $X=BA^{-1}$ — the side on which $A^{-1}$ is applied must match the side on which $A$ originally …