Question 25: Without using row reduction, find the inverse of \(A = \left[ {\begin{array}{*{20}{c}}1&2&0&0&0\\3&5&0&0&0\\0&0&2&0&0\\0&0&0&7&8\\0&0&0&5&6\end{array}} \right]\).

Consider the partition of A as the\(2 \times 2\) block diagonal matrix shown below.So, the \(\left( {2,2} \right)\)-block is a block-diagonal matrix.

\(\begin{gathered}A = \left[ {\begin{array}{*{20}{c}} 1&2& & 0&0&0 \\ 3&5& & 0&0&0 \\ \hline 0&0& & 2&0&0 \\ 0&0& & 0&7&8 \\ 0&0& & 0&5&6 \end{array}} \right] \\ = \left[ {\begin{array}{*{20}{c}} {{A_{11}}}& & 0 \\ \hline 0&& {{A_{22}}} \end{array}} \right] \\ \end{gathered} \)

Here,{A_{22}} = \(\left[ {\begin{array}{*{20}{c}}2& & 0&0 \\ \hline0& & 7&8 \\ 0& & 5&6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&0 \\ 0&B \end{array}} \right]\)

.

It is observed that \(B\) is invertible, and the inverse of Bis shown below.

\(\begin{array}{c}{B^{ - 1}} = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}6&{ - 8}\\{ - 5}&7\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}3&{ - 4}\\{ - 2.5}&{3.5}\end{array}} \right]\end{array}\)

Exercise 13 shows that the block diagonal matrix \({A_{22}}\) is invertible.

It is observed that \({A_{11}}\) is also invertible.The inverse of \({A_{11}}\) is shown below.

\(\begin{array}{c}{A_{11}} = \frac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}5&{ - 2}\\{ - 3}&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}&2\\3&{ - 1}\end{array}} \right]\end{array}\)

From exercise 13,\(A\)itself is invertible, and the inverse of Ais block diagonal.

\(\begin{gathered} {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {A_{11}^{ - 1}}&0 \\ 0&{A_{22}^{ - 1}} \end{array}} \right] \\ = \left[ {\begin{array}{*{20}{c}}{ - 5}&2& & {}&{}&{} \\ 3&{ - 1}& & {}&{\begin{array}{*{20}{c}}0 \\ {} \end{array}}&{} \\ \hline{}&{}& & {.5}&0&0 \\ {}&{\begin{array}{*{20}{c}}0&{} \end{array}}& & 0&3&{ - 4} \\ {}&{}& & 0&{ - 2.5}&{3.5} \end{array}} \right] \\ = \left[ {\begin{array}{*{20}{c}} { - 5}&2&0&0&0 \\ 3&{ - 1}&0&0&0 \\ 0&0&{.5}&0&0 \\ 0&0&0&3&{ - 4} \\ 0&0&0&{ - 2.5}&{3.5} \end{array}} \right] \\ \end{gathered}\)

Thus, the inverse of Ais \({A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}{ - 5}&2&0&0&0\\3&{ - 1}&0&0&0\\0&0&{.5}&0&0\\0&0&0&3&{ - 4}\\0&0&0&{ - 2.5}&{3.5}\end{array}} \right]\).