# What is the correct option?

## #### Answer:

Option (D) $9 I - A$ is correct

#### Explanation:

Find ${A}^{- 1} = \setminus \frac{\setminus a \mathrm{dj} \left(A\right)}{| A |}$

Check the options. Option (D) satisfies the given condition

Jul 28, 2018

#### Answer:

The answer is $o p t i o n \left(D\right)$

#### Explanation:

The matrix is

$A = \left(\begin{matrix}2 & - 3 \\ - 4 & 7\end{matrix}\right)$

The inverse of a matrix

$A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ is

${A}^{-} 1 = \frac{1}{\det A} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$

The determinant of

$| A | = a d - b c = 2 \cdot 7 - \left(- 3 \cdot - 4\right) = 14 - 12 = 2$

Therefore,

${A}^{-} 1 = \frac{1}{2} \left(\begin{matrix}7 & 3 \\ 4 & 2\end{matrix}\right)$

And

$2 {A}^{-} 1 = \left(\begin{matrix}7 & 3 \\ 4 & 2\end{matrix}\right)$

$9 I - A = 9 \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) - \left(\begin{matrix}2 & - 3 \\ - 4 & 7\end{matrix}\right)$

$= \left(\begin{matrix}7 & 3 \\ 4 & 2\end{matrix}\right)$

$= 2 {A}^{-} 1$

The answer is $o p t i o n \left(D\right)$