Let A = [ * 20 c 1&2&2 2&1&2 2&2&1 ] , then - Vidyadip

MCQ

Let $A =$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ , then

A
$A^2 - 4A - 5I_3 = 0$
B
$A^{-1} = \frac{1}{5} (A - 4I_3)$
C
$A^2$ is invertible
✓
All of the above

✓

Answer

Correct option: D.

All of the above

d
$A^2 =$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $=$ $\left[ {\begin{array}{*{20}{c}}9&8&8\\8&9&8\\8&8&9\end{array}} \right]$
We have $A^2 - 4A - 5I_3$

$=$ $\left[ {\begin{array}{*{20}{c}}9&8&8\\8&9&8\\8&8&9\end{array}} \right]$ $- 4$ $ \left[ {\begin{array}{*{20}{c}}1&2&2\\2&1&2\\2&2&1\end{array}}\right]$ $- 5$ $\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$ $= O$

==> $5I_3 = A^2 - 4A = A(A - 4I_3)$

==>$I_3 = A$ $\left[ {\frac{1}{5}(A - 4{I_3})} \right]$

==> $A^{-1} =$ $\frac{1}{5}(A - 4{I_3})$

Note that $|A| = 5$. Since $|A^3| = |A|^3 = 5^3 \ne 0, A^3$ is invertibleSimilarly, $A^2$ is invertible

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