SATHEE: Matrices and Determinants 1 Question 2

Matrices and Determinants 1 Question 2

2. The total number of matrices $A = [\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & - 1 \\ 2 x & - y & 1 \end{array}]$ , $(x, y \in R, x \neq y)$ for which $A^{T} A = 3 I_{3}$ is

(2019 Main, 9 April II)

(a) 2

(b) 4

(d) 6

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Answer:

Correct Answer: 2. (b)

Solution:

Given matrix

$A = [\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & - 1 \\ 2 x & - y & 1 \end{array}], (x, y \in R, x \neq y)$

for which

$\begin{aligned} A^{T} A = 3 I_{3} \\ \Rightarrow & [\begin{array}{ccc} 0 & 2 x & 2 x \\ 2 y & y & - y \\ 1 & - 1 & 1 \end{array}] [\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & - 1 \\ 2 x & - y & 1 \end{array}] = [\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}] \\ \Rightarrow & [\begin{array}{ccc} 8 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 \end{array}] = [\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}] \end{aligned}$

Here, two matrices are equal, therefore equating the corresponding elements, we get

$\begin{aligned} 8 x^{2} & = 3 and 6 y^{2} = 3 \\ \Rightarrow & x = \pm \sqrt{\frac{3}{8}} \\ and & y = \pm \frac{1}{\sqrt{2}} \end{aligned}$

$∵$ There are 2 different values of $x$ and $y$ each.

So, 4 matrices are possible such that $A^{T} A = 3 I_{3}$ .

Matrices and Determinants 1 Question 2

2. The total number of matrices $A = [\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & - 1 \\ 2 x & - y & 1 \end{array}]$ , $(x, y \in R, x \neq y)$ for which $A^{T} A = 3 I_{3}$ is

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Matrices and Determinants 1 Question 2

2. The total number of matrices A=[02y12xy−12x−y1], (x,y∈R,x≠y) for which ATA=3I3 is

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2. The total number of matrices $A = [\begin{array}{ccc} 0 & 2 y & 1 \\ 2 x & y & - 1 \\ 2 x & - y & 1 \end{array}]$ , $(x, y \in R, x \neq y)$ for which $A^{T} A = 3 I_{3}$ is