Matrices And Determinants Ques 22
The total number of matrices $A=\left[\begin{array}{ccc}0 & 2 y & 1 \\ 2 x & y & -1 \\ 2 x & -y & 1\end{array}\right]$, $(x, y \in R, x \neq y)$ for which $A^T A=3 I_3$ is
(2019 Main, 9 April II)
(a) $2$
(b) $4$
(c) $3$
(d) $6$
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Answer:
Correct Answer: 22.(b)
Solution:
Formula:
Properties of matrix multiplication:
Given matrix
$ A=\left[\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & -1 \\ 2 x & -y & 1 \end{array}\right],(x, y \in R, x \neq y) $
for which
$ \begin{aligned} & A^T A=3 I_3 \\ \Rightarrow & {\left[\begin{array}{ccc} 0 & 2 x & 2 x \\ 2 y & y & -y \\ 1 & -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & -1 \\ 2 x & -y & 1 \end{array}\right]=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right] } \\ \Rightarrow & {\left[\begin{array}{ccc} 8 x^2 & 0 & 0 \\ 0 & 6 y^2 & 0 \\ 0 & 0 & 3 \end{array}\right]=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right] } \end{aligned} $
Here, two matrices are equal, therefore equating the corresponding elements, we get
$ 8 x^2 =3 \text { and } 6 y^2=3 $
$\Rightarrow x = \pm \sqrt{\frac{3}{8}} $
$\text { and } y = \pm \frac{1}{\sqrt{2}}$
$\because$ There are $2$ different values of $x$ and $y$ each.
So, $ 4 $ matrices are possible such that $A^T A=3 I_3$.
BETA
