Matrices and Determinants 2 Question 21

23. 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

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Answer:

Correct Answer: 23. $(b,c)$

Solution:

  1. 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 $



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