Question

Let $A=\begin{bmatrix} -4 & -3 & -3 \\ 1 & a & 1 \\ 4 & b & 3 \end{bmatrix}$ and $A=A^{- 1}$ , then $a+2b$ is equal to

• A. $0$
• B. $4$
• C. $8$
• D. $5$

Answer 1

Answer: (C)

$\therefore A=A^{- 1}\Rightarrow A^{2}=I$
$A^{2} = A \cdot A = \begin{bmatrix} - 4 & - 3 & - 3 \\ 1 & a & 1 \\ 4 & b & 3 \end{bmatrix} \begin{bmatrix} - 4 & - 3 & - 3 \\ 1 & a & 1 \\ 4 & b & 3 \end{bmatrix}$
$=\begin{bmatrix} 1 & 12-3a-3b & 0 \\ a & -3+a^{2}+b & a \\ b-4 & -12+ab+3b & b-3 \end{bmatrix}$
$\therefore A^{2}=I\Rightarrow a=0,b=4\Rightarrow a+2b=8$