3 and 4 .Determinants and Matrices
easy

If $A = \left[ {\begin{array}{*{20}{c}}4&1\\3&2\end{array}} \right]$and $I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$, ${A^2} - 6A = $

A

$3I$

B

$5I$

C

$-5I$

D

None of these

Solution

(c) ${A^2} – 6\,A = \left[ {\begin{array}{*{20}{c}}4&1\\3&2\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}4&1\\3&2\end{array}} \right] – 6\left[ {\begin{array}{*{20}{c}}4&1\\3&2\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{19}&6\\{18}&7\end{array}} \right] – \left[ {\begin{array}{*{20}{c}}{24}&6\\{18}&{12}\end{array}} \right]$$\left[ {\begin{array}{*{20}{c}}{ – 5}&0\\0&{ – 5}\end{array}} \right] = – 5I$.

Std 12
Mathematics

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