Question

Show that the matrix is nilpotent matrix of order 3

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Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm :\longmapsto\:A = \begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5& 2&6\\ - 2& - 1& - 3\end{array}\right]\end{gathered}$

Now, Consider

$\rm :\longmapsto\: {A}^{2}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5& 2&6\\ - 2& - 1& - 3\end{array}\right]\end{gathered} \times \begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5& 2&6\\ - 2& - 1& - 3\end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}1 + 5 - 6&1 + 2 - 3&3 + 6 - 9\\5 + 10 - 12& 5 + 4 - 6&15 + 1218\\ - 2 - 5 + 6& - 2 - 2 + 3& - 6 - 6 + 9\end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\3&3&9\\ - 1& - 1& - 3\end{array}\right]\end{gathered}$

Now, Consider

$\rm :\longmapsto\: {A}^{3}$

$\rm \:  =  \: {A}^{2} \times A$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\3&3&9\\ - 1& - 1& - 3\end{array}\right]\end{gathered} \times \begin{gathered}\sf \left[\begin{array}{ccc}1&1&3\\5& 2&6\\ - 2& - 1& - 3\end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0 + 0 + 0&0 + 0 + 0&0 + 0 + 0\\3 + 15 - 18& 3 + 6 - 9&9 + 18 - 27\\ - 1 - 5 + 6& - 1 - 2 + 3& - 3 - 6 + \end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf \left[\begin{array}{ccc}0&0&0\\0& 0&0\\ 0&0&0\end{array}\right]\end{gathered}$

$\bf\implies \: {A}^{3} = 0$

$\bf\implies \:A \: is \: nilpotent \: matrix\: of \: order \: 3$

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Different types of Matrices

$\begin{gathered}\boxed{\begin{array}{c|c} \bf & \bf \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf Idempotent \: Matrix & \sf {A}^{2} = A \\ \\ \sf Nilpotent \: Matrix & \sf {A}^{k} = 0 \\ \\ \sf Periodic \: Matrix & \sf {A}^{k + 1} = A\\ \\ \sf Involutory \: Matrix & \sf {A}^{2} =I \end{array}} \\ \end{gathered}$