$\operatorname{adj}\left[\begin{array}{ccc}1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b\end{array}\right], \quad$ then $\left[\begin{array}{ll}a & b\end{array}\right]$ is equal to

• A

  [− 4 1]
• B

  [− 4 −1]
• Correct answer
  C

  [4 1]
• D

  [4 −1]

Solution

Given that
$\begin{aligned} & \text { Cofactor of }\left[\begin{array}{ccc} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{array}\right] \text { are } \\ & C_{11}=5, C_{12}=1, C_{13}=-2 \\ & C_{21}=4, C_{22}=1, C_{23}=-2 \\ & C_{31}=-2, C_{32}=0, C_{33}=1 \\ & \Rightarrow\left[\begin{array}{ccc} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{array}\right]=\left[\begin{array}{ccc} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{array}\right] \end{aligned}$ On comparing the corresponding elements, we get $\begin{array}{rlrl} & a & =4, b=1 \\ \therefore & {\left[\begin{array}{ll} a & b \end{array}\right]} & =\left[\begin{array}{ll} 4 & 1 \end{array}\right] \end{array}$

Hence, the correct answer is option: 'C'.