MCQ
If $A = \left[ {\begin{array}{*{20}{c}}
3&7\\
1&2
\end{array}} \right]$ then $|A^{2011} -5A^{2010}|$ is equal to
3&7\\
1&2
\end{array}} \right]$ then $|A^{2011} -5A^{2010}|$ is equal to
- A$1$
- ✓$-1$
- C$6$
- D$-6$
✓
Answer
Correct option: B.
$-1$
b
$\left|\mathrm{A}^{2011}-5 \mathrm{A}^{2010}\right|=|\mathrm{A}|^{2010}|\mathrm{A}-5 \mathrm{I}|$
$\left|\mathrm{A}^{2011}-5 \mathrm{A}^{2010}\right|=|\mathrm{A}|^{2010}|\mathrm{A}-5 \mathrm{I}|$
$A-5 I=\left(\begin{array}{ll}{3} & {7} \\ {1} & {2}\end{array}\right)-\left(\begin{array}{ll}{5} & {0} \\ {0} & {5}\end{array}\right)=\left(\begin{array}{cc}{-2} & {7} \\ {1} & {-3}\end{array}\right)$
$|A-5 I|=-1$ and $|A|=-1$
So $|\mathrm{A}|^{2010}|\mathrm{A}-5 \mathrm{I}|=(-1)^{2010}(-1)=-1$
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