Subjective Type

If A=[ 1 2 2 2 1 2 2 2 1 ], then show that A2−4A−5I=0, where I and 0 are the unit matrix and the null matrix of order 3 , respectively. Use this result to find A−1.

Solution

Given $$A=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] $$

$${ A }^{ 2 }=\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] $$

$$\therefore { A }^{ 2 }=\left[ \begin{matrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{matrix} \right] $$

$$\therefore { A }^{ 2 }=\left[ \begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix} \right] $$

$$\therefore { A }^{ 2 }-4A-5I=\left[ \begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix} \right] -4\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] -5\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] $$

$$=\left[ \begin{matrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{matrix} \right] -\left[ \begin{matrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{matrix} \right] -\left[ \begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix} \right] $$

$$=\left[ \begin{matrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{matrix} \right] $$

$$=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] $$

$$\therefore { A }^{ 2 }-4A-5I=0$$

Multiply both sides by $${ A }^{ -1 }$$, we get,

$${ A }-4-5I.{ A }^{ -1 }=0$$

$$\therefore { A }-4-5{ A }^{ -1 }=0$$

$$\therefore 5{ A }^{ -1 }={ A }-4I$$

$$\therefore { A }^{ -1 }=\frac { 1 }{ 5 } \left( { A }-4I \right) $$

$$\therefore { A }^{ -1 }=\frac { 1 }{ 5 } \left\{ \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] -4\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \right\} $$

$$\therefore { A }^{ -1 }=\frac { 1 }{ 5 } \left\{ \left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \right\} $$

$$\therefore { A }^{ -1 }=\frac { 1 }{ 5 } \left[ \begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix} \right] $$

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