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The matrix $A^2 + 4A - 5I$, where $I$ is identity matrix and $A = \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ - 3}
\end{array}} \right]$ , equals
$4\left[ {\begin{array}{*{20}{c}}
2&1\\
2&0
\end{array}} \right]$
$4\left[ {\begin{array}{*{20}{c}}
0&{ - 1}\\
2&2
\end{array}} \right]$
$32\left[ {\begin{array}{*{20}{c}}
2&1\\
2&0
\end{array}} \right]$
$32\left[ {\begin{array}{*{20}{c}}
1&1\\
1&0
\end{array}} \right]$
Solution
${A^2} + 4A – 51 = A \times A + 4A – 5I$
$ = \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ – 3}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ – 3}
\end{array}} \right] + 4\left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ – 3}
\end{array}} \right] – 5\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
9&{ – 4}\\
{ – 8}&{17}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
4&8\\
{16}&{ – 12}
\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}
5&0\\
0&5
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{9 + 4 – 5}&{ – 4 + 8 – 0}\\
{ – 8 + 16 – 0}&{17 – 12 – 5}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
8&4\\
8&0
\end{array}} \right]$
$ = 4\left[ {\begin{array}{*{20}{c}}
2&1\\
2&0
\end{array}} \right]$