Matrices And Determinants Ques 45
Which of the following values of $\alpha$ satisfy the equation
$\begin{vmatrix}(1+\alpha)^{2} & (1+2 \alpha)^{2} & (1+3 \alpha)^{2} \\ (2+\alpha)^{2} & (2+2 \alpha)^{2} & (2+3 \alpha)^{2} \\ (3+\alpha)^{2} & (3+2 \alpha)^{2} & (3+3 \alpha)^{2}\end{vmatrix}=-648 \alpha$ ?
(a) $-4$
(b) $9$
(c) $-9$
(d) $4$
2015 Adv.
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Answer:
Correct Answer: 45.$(b,c)$
Solution:
Formula:
Properties of matrix multiplication:
- Given determinant could be expressed as product of two determinants.
$ \text { i.e. }\begin{vmatrix} (1+\alpha)^{2} & (1+2 \alpha)^{2} & (1+3 \alpha)^{2} \\ (2+\alpha)^{2} & (2+2 \alpha)^{2} & (2+3 \alpha)^{2} \\ (3+\alpha)^{2} & (3+2 \alpha)^{2} & (3+3 \alpha)^{2} \end{vmatrix}=-648 \alpha $
$ \Rightarrow\begin{vmatrix} 1+2 \alpha+\alpha^{2} & 1+4 \alpha+4 \alpha^{2} & 1+6 \alpha+9 \alpha^{2} \\ 4+4 \alpha+\alpha^{2} & 4+8 \alpha+4 \alpha^{2} & 4+12 \alpha+9 \alpha^{2} \\ 9+6 \alpha+\alpha^{2} & 9+12 \alpha+4 \alpha^{2} & 9+18 \alpha+9 \alpha^{2} \end{vmatrix} $ $-648 \alpha$
$\Rightarrow \quad \begin{vmatrix} 1 & \alpha & \alpha^{2} \\ 4 & 2 \alpha & \alpha^{2} \\ 9 & 3 \alpha & \alpha^{2} \end{vmatrix} \cdot\begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & 6 \\ 1 & 4 & 9 \end{vmatrix}=-648 \alpha $
$\Rightarrow \quad \alpha^{3}\begin{vmatrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{vmatrix} \cdot\begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & 6 \\ 1 & 4 & 9 \end{vmatrix}=-648 \alpha $
$ \Rightarrow \quad -8 \alpha^{3} =-648 \alpha $
$\Rightarrow \quad \alpha^{3}-81 \alpha=0 \Rightarrow \quad \alpha\left(\alpha^{2}-81\right)=0$
$ \therefore \quad \alpha= 0, \pm 9 $
BETA
