Q85 / 180 · KCET 2024 · Mathematics

If $P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix $A$ and $|A|=4$, then $\alpha$ is equal to

  • A
    $4$
  • B
    $5$
  • Correct answer
    C
    $11$
  • D
    $0$

Solution

$P=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$ For $3 \times 3$, matrix, $|\operatorname{adj} A|=|A|^2$ $|\operatorname{adj} A|=(4)^2=16$ Now, $\quad P=\left|\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right|=16$ $\Rightarrow \quad 1(12-12)-\alpha(4-6)+3(4-6)=16$ $\Rightarrow \quad 0+2 \alpha-6=16$ $\Rightarrow \quad 2 \alpha=22$ Hence, $\alpha=11$

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

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