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
$=$ $\left[ {\begin{array}{*{20}{c}}9&8&8\\8&9&8\\8&8&9\end{array}} \right]$ $- 4$ $ \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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| Column | Maximum of $z$ |
| $A$ | $300$ |
| $B$ | $325$ |