Prove the following

Question:

If $\mathrm{A}=\left(\begin{array}{ll}2 & 2 \\ 9 & 4\end{array}\right)$ and $\mathrm{I}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$, then $10 \mathrm{~A}^{-1}$ is equal to:

  1. (1) $A-4 I$

  2. (2) $6 I-A$

  3. (3) $A-6 I$

  4. (4) $4 I-A$


Correct Option: , 3

Solution:

Characteristics equation of matrix ' $A$ ' is $|A-\lambda I|=0$

$\left|\begin{array}{cc}2-\lambda & 2 \\ 9 & 4-\lambda\end{array}\right|=0 \Rightarrow \lambda^{2}-6 \lambda-10=0$

$\therefore \quad A^{2}-6 A-10 I=0$

$\Rightarrow A^{-1}\left(A^{2}\right)-6 A^{-1}-10 I A^{-1}=0$

$\Rightarrow 10 A^{-1}=A-6 I$

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