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यदि $A = \left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&{ - 1}\\3&{ - 1}&1\end{array}} \right]$, तो
${A^3} + 3{A^2} + A - 9{I_3} = O$
${A^3} - 3{A^2} + A + 9{I_3} = O$
${A^3} + 3{A^2} - A + 9{I_3} = O$
${A^3} - 3{A^2} - A + 9{I_3} = O$
Solution
(d) $A = \left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&{ – 1}\\3&{ – 1}&1\end{array}} \right]$
${A^2} = A\,.\,A = \left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&{ – 1}\\3&{ – 1}&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&{ – 1}\\3&{ – 1}&1\end{array}} \right]\,$$ = \left[ {\begin{array}{*{20}{c}}4&3&0\\{ – 3}&2&{ – 2}\\6&4&5\end{array}} \right]$
$A\,.\,{A^2} = \left[ {\begin{array}{*{20}{c}}1&2&1\\0&1&{ – 1}\\3&{ – 1}&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}4&3&0\\{ – 3}&2&{ – 2}\\6&4&5\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}4&{11}&1\\{ – 9}&{ – 2}&{ – 7}\\{21}&{11}&7\end{array}} \right]$
==> ${A^3} – 3{A^2} – A + 9{I_3} = 0$.