Question
Show that the matrix $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ satisfies the equation $A ^2-4 A + I =0$. where $I$ is $2 \times 2$ identity matrix and $O$ is $2 \times 2$ zero matrix. Using this equation, find $A^{-1}$
$= \left[ {\begin{array}{*{20}{c}} 7&{12} \\ 1&7 \end{array}} \right]$
${A^2} - 4A + I = \left[ {\begin{array}{*{20}{c}} 7&{12} \\ 1&7 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 8&{12} \\ 4&8 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
= 0
${A^2} - 4A + I = 0$
${A^2} - 4A = - I$
$AA{A^{ - 1}} - 4A{A^{ - 1}} = I{A^{ - 1}}$
$AI - 4I = - I{A^{ - 1}}\left[ {\because A{A^{ - 1}} = I} \right]$
${A^{ - 1}} = 4I - A$
$= \left[ {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 1}&2 \end{array}} \right]$
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