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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 $\beta$ be a real number. Consider the matrix

$A=\left(\begin{array}{ccc}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{array}\right)$

If $A^7-(\beta-1) A^6-\beta A^5$ is a singular matrix, then the value of $9 \beta$ is

  • A
    $2$
  • ✓
    $3$
  • C
    $4$
  • D
    $5$
✓

Answer

Correct option: B.
$3$
b
$A=\left(\begin{array}{ccc}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{array}\right)|A|=-1$

$\Rightarrow\left|A^7-(\beta-1) A^6-\beta A^5\right|=0$

$\Rightarrow|A|^5\left|A^2-(\beta-1) A-\beta I\right|=0$

$\Rightarrow|A|^5\left|\left(A^2-\beta A\right)+A-\beta I\right|=0$

$\Rightarrow|A|^5|A(A-\beta I)+I(A-\beta I)|=0$

$|A|^5|(A+I)(A-\beta I)|=0$

$\begin{array}{l}A+I=\left(\begin{array}{ccc}\beta+1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1\end{array}\right) \Rightarrow|A+I|=-4, \text { Here }|A| \neq 0 \&|A+I| \neq 0 \\ A-\beta I=\left(\begin{array}{ccc}0 & 0 & 1 \\ 2 & 1-\beta & -2 \\ 3 & 1 & -2-\beta\end{array}\right)\end{array}$

$\begin{array}{l}|A-\beta I|=2-3(1-\beta)=3 \beta-1=0 \Rightarrow \beta=\frac{1}{3} \\ 9 \beta=3\end{array}$

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