Question

If $A=\left[\begin{array}{cc}2& 2\\ 9& 4\end{array}\right]$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ then $10A^{-1}$ is equal to:

• A. $A+6I$
• B. $A-6I$
• C. $A+4I$
• D. $A-4I$

Answer 1

The correct option is B

$A-6I$

Explanation for the correct option:

Inverse of a matrix:

The determinant of $A$ is given as

$\Rightarrow$$\left|A\right|=\left|\begin{array}{cc}2& 2\\ 9& 4\end{array}\right|$

$\Rightarrow$$\left|A\right|=2\times 4-2\times 9$ $\left[\because A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\Rightarrow \left|A\right|=\mathrm{ad}-\mathrm{bc}\right]$

$\Rightarrow$$\left|A\right|=-10$

The co-factor matrix of $A$ is given as

$\Rightarrow$$A_{cf}=\left[\begin{array}{cc}4{\left(-1\right)}^{1+1}& 9{\left(-1\right)}^{1+2}\\ 2{\left(-1\right)}^{2+1}& 2{\left(-1\right)}^{2+2}\end{array}\right]$ $\left[\because A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\Rightarrow A_{\mathrm{cf}}=\left[\begin{array}{cc}d{\left(-1\right)}^{1+1}& c{\left(-1\right)}^{1+2}\\ b{\left(-1\right)}^{2+1}& a{\left(-1\right)}^{2+2}\end{array}\right]\right]$

$\Rightarrow$$A_{cf}=\left[\begin{array}{cc}4& -9\\ -2& 2\end{array}\right]$

The adjoint of $A$ is the transpose of the co factor matrix

$\Rightarrow$$adjA={\left(A_{cf}\right)}^T$

$\Rightarrow$$adjA=\left[\begin{array}{cc}4& -2\\ -9& 2\end{array}\right]$

The inverse of $A$ is given by

$\Rightarrow$$A^{-1}=\frac{adjA}{\left|A\right|}$

$\Rightarrow$$A^{-1}=\frac{-1}{10}\left[\begin{array}{cc}4& -2\\ -9& 2\end{array}\right]$

$\Rightarrow$$A^{-1}=\frac{1}{10}\left[\begin{array}{cc}-4& 2\\ 9& -2\end{array}\right]$

$\Rightarrow$$10A^{-1}=\left[\begin{array}{cc}-4& 2\\ 9& -2\end{array}\right]$ $...\left(i\right)$

Now we find $A-6I$

$\Rightarrow$$A-6I=\left[\begin{array}{cc}2& 2\\ 9& 4\end{array}\right]-\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]$

$\Rightarrow$$A-6I=\left[\begin{array}{cc}-4& 2\\ 9& -2\end{array}\right]$ $...\left(ii\right)$

From $\left(i\right),\left(ii\right)$

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

Hence option (B) is the correct answer.

Explanation for the in-correct option:

Option (A):$A+6I$

$A+6I=\left[\begin{array}{cc}2& 2\\ 9& 4\end{array}\right]+\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]$

$\Rightarrow$$A+6I=\left[\begin{array}{cc}8& 2\\ 9& 10\end{array}\right]$

Which is not equal to $10A^{-1}$.

Option (C):$A+4I$

$A+4I=\left[\begin{array}{cc}2& 2\\ 9& 4\end{array}\right]+\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]$

$\Rightarrow$$A+4I=\left[\begin{array}{cc}6& 2\\ 9& 8\end{array}\right]$

Which is not equal to $10A^{-1}$.

Option (D):$A-4I$

$A-4I=\left[\begin{array}{cc}2& 2\\ 9& 4\end{array}\right]-\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]$

$\Rightarrow$$A-4I=\left[\begin{array}{cc}-2& 2\\ 9& 0\end{array}\right]$

Which is not equal to $10A^{-1}$.

Hence option (B) is the correct answer.