Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

Let the inverse of matrix\(A\) be

\({A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}D&E\\F&G\end{array}} \right]\).

Then,

\(\begin{array}{c}A{A^{ - 1}} = I\\\left[ {\begin{array}{*{20}{c}}B&0\\0&C\end{array}} \right]\left[ {\begin{array}{*{20}{c}}D&E\\F&G\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{BD}&{BE}\\{CF}&{CG}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right].\end{array}\)

Since \(B\) is a square and \(BD = I\), \(B\) is invertible.Similar is the case for \(C\).

So, \({A^{ - 1}}\) can be expressed as

\(\begin{array}{c}{A^{ - 1}} = {\left[ {\begin{array}{*{20}{c}}B&0\\0&C\end{array}} \right]^{ - 1}}\\ = \left[ {\begin{array}{*{20}{c}}{{B^{ - 1}}}&0\\0&{{C^{ - 1}}}\end{array}} \right].\end{array}\)

Hence, it is proved that \(A\) is invertible if and only if \(B\) and \(C\) are invertible.