Question

a. Verify that \({A^2} = I\) when \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\).

b. Use partitioned matrices to show that \({M^2} = I\) when\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\).

Short answer

1. It is verified that \({A^2} = I\).
2. It is proved that \({M^2} = I\).

Step-by-step solution

Verify that \({A^2} = I\)

(a)

If \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\),

\(\begin{array}{c}{A^2} = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + 0}&{0 + 0}\\{3 - 3}&{0 + {{\left( { - 1} \right)}^2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = I.\end{array}\)

Thus, it is verified that \({A^2} = I\).

Determine the partitioned matrix of M

(b)

Matrix \(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\)can be written as a \(2 \times 2\) partitioned matrix.

\(\begin{array}{c}M = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\end{array}\)

Use the partitioned matrix to show that \({M^2} = I\)

If \(M = \left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\),

\[\begin{array}{c}{M^2} = \left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&0\\I&{ - A}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{A^2} + 0}&{0 + 0}\\{A - A}&{0 + {{\left( { - A} \right)}^2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]\,\,\,\,\,\,{\mathop{\rm since}\nolimits} \,\,\,{A^2} = I\\ = I.\end{array}\]

Thus, it is proved that \({M^2} = I\).