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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
Let $B=\left[\begin{array}{ccc}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha > 2$ be the adjoint of $a$ matrix $A$ and $| A |=2$, then $[\alpha\,\,-2 \alpha \,\, \alpha \,\,] B \left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right]$ is equal to :-
  • A
    $16$
  • B
    $32$
  • ✓
    $-16$
  • D
    $0$
✓

Answer

Correct option: C.
$-16$
c
Given, $B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right]$

$|B|=4$

$1(8-3 \alpha)-3(4-3 \alpha)+\alpha(\alpha-2 \alpha)=4$

$-\alpha^2+6 \alpha-8=0$

$\alpha=2,4$

Given,$\alpha > 2$

So,$\alpha=2$ is rejected

$\left[\begin{array}{lll}4 & -8 & 4\end{array}\right]\left[\begin{array}{lll}1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4\end{array}\right]\left[\begin{array}{c}4 \\ -8 \\ 4\end{array}\right]=[-16]_{1 \times 1}$

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