Q247 / 270 · NDA 2017 (Phase 2) · Mathematics

The adjoint of the matrix $A=\left[\begin{array}{lll}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{array}\right]$ is

  • A
    $\left[\begin{array}{lrr}-1 & 6 & 2 \\ -2 & 1 & -4 \\ 6 & 3 & 1\end{array}\right]$
  • Correct answer
    B
    $\left[\begin{array}{lcr}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]$
  • C
    $\left[\begin{array}{rrr}6 & 1 & 2 \\ 4 & -1 & 2 \\ 6 & 3 & -1\end{array}\right]$
  • D
    $\left[\begin{array}{lll}-6 & 2 & 1 \\ 4 & -2 & 1 \\ 3 & 1 & -6\end{array}\right]$

Solution

$\begin{aligned} & A=\left[\begin{array}{lll}1 & 0 & 2 \\ 2 & 1 & 0 \\ 0 & 3 & 1\end{array}\right] \\ & A_{11}=1-0=1 ; A_{12}=-(2-0)=-2 ; A_{13}=6-0=6 \\ & A_{21}=-(0-6)=6 ; A_{22}=1-0=1 ; A_{23}=-3-0=-3 \\ & A_{31}=0-2=-2 ; A_{32}=-(0-4)=4 ; A_{33}=1-0=1\end{aligned}$ $\therefore \operatorname{Adj} A=\left[\begin{array}{ccc}1 & 6 & -2 \\ -2 & 1 & 4 \\ 6 & -3 & 1\end{array}\right]$

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

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