The total number of matrices A = ( \beginmatrix 0 & 2y & 1 2x & y & - 1 2x & - y & 1 \endmatrix ) (x, y \in R,x e y) for which ATA = 3I3 is :-

Matrices and Determinants PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q36 with Solution

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Correct answer: B

We are given

A=\begin{pmatrix} 0 & 2y & 1\\ 2x & y & -1\\ 2x & -y & 1 \end{pmatrix}

and need the number of real pairs $(x,y)$ for which $A^T A = 3I_3.$

Recall: if $A^T A = 3I_3$ , then the columns of $A$ must be pairwise orthogonal, and each column must have squared length $3$ .

Let the columns of $A$ be:

c_1=\begin{pmatrix}0\\2x\\2x\end{pmatrix}, \quad c_2=\begin{pmatrix}2y\\y\\-y\end{pmatrix}, \quad c_3=\begin{pmatrix}1\\-1\\1\end{pmatrix}.

Then

c_1\cdot c_1 & c_1\cdot c_2 & c_1\cdot c_3\\ c_2\cdot c_1 & c_2\cdot c_2 & c_2\cdot c_3\\ c_3\cdot c_1 & c_3\cdot c_2 & c_3\cdot c_3 \end{pmatrix}.

Compute the dot products.

First column with itself:

c_1\cdot c_1 = 0^2+(2x)^2+(2x)^2 = 8x^2.

Since this must equal $3$ ,

8x^2=3 \implies x^2=\frac{3}{8} \implies x=\pm \sqrt{\frac{3}{8}}.

Second column with itself:

c_2\cdot c_2 = (2y)^2+y^2+(-y)^2 = 6y^2.

Since this must equal $3$ ,

6y^2=3 \implies y^2=\frac12 \implies y=\pm \frac{1}{\sqrt2}.

Third column with itself:

c_3\cdot c_3 = 1^2+(-1)^2+1^2=3,

which already matches.

Now check orthogonality.

$c_1\cdot c_2$ :

c_1\cdot c_2 = 0(2y)+(2x)(y)+(2x)(-y)=2xy-2xy=0.

So this is always satisfied.

$c_1\cdot c_3$ :

c_1\cdot c_3 = 0(1)+(2x)(-1)+(2x)(1) = -2x+2x=0.

Always satisfied.

$c_2\cdot c_3$ :

c_2\cdot c_3 = (2y)(1)+y(-1)+(-y)(1)=2y-y-y=0.

Also always satisfied.

Thus the only conditions are:

x=\pm \sqrt{\frac{3}{8}}, \qquad y=\pm \frac{1}{\sqrt2}.

Number of possible matrices:

$x$ has $2$ choices,
$y$ has $2$ choices.

Hence total number of matrices is

2\times 2 = 4.

Therefore the correct option is $\boxed{4}$ which is option B.

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View written solutionFree

Correct answer: B

We are given

A=\begin{pmatrix} 0 & 2y & 1\\ 2x & y & -1\\ 2x & -y & 1 \end{pmatrix}

and need the number of real pairs $(x,y)$ for which $A^T A = 3I_3.$

Recall: if $A^T A = 3I_3$ , then the columns of $A$ must be pairwise orthogonal, and each column must have squared length $3$ .

Let the columns of $A$ be:

c_1=\begin{pmatrix}0\\2x\\2x\end{pmatrix}, \quad c_2=\begin{pmatrix}2y\\y\\-y\end{pmatrix}, \quad c_3=\begin{pmatrix}1\\-1\\1\end{pmatrix}.

Then

c_1\cdot c_1 & c_1\cdot c_2 & c_1\cdot c_3\\ c_2\cdot c_1 & c_2\cdot c_2 & c_2\cdot c_3\\ c_3\cdot c_1 & c_3\cdot c_2 & c_3\cdot c_3 \end{pmatrix}.

Compute the dot products.

First column with itself:

c_1\cdot c_1 = 0^2+(2x)^2+(2x)^2 = 8x^2.

Since this must equal $3$ ,

8x^2=3 \implies x^2=\frac{3}{8} \implies x=\pm \sqrt{\frac{3}{8}}.

Second column with itself:

c_2\cdot c_2 = (2y)^2+y^2+(-y)^2 = 6y^2.

Since this must equal $3$ ,

6y^2=3 \implies y^2=\frac12 \implies y=\pm \frac{1}{\sqrt2}.

Third column with itself:

c_3\cdot c_3 = 1^2+(-1)^2+1^2=3,

which already matches.

Now check orthogonality.

$c_1\cdot c_2$ :

c_1\cdot c_2 = 0(2y)+(2x)(y)+(2x)(-y)=2xy-2xy=0.

So this is always satisfied.

$c_1\cdot c_3$ :

c_1\cdot c_3 = 0(1)+(2x)(-1)+(2x)(1) = -2x+2x=0.

Always satisfied.

$c_2\cdot c_3$ :

c_2\cdot c_3 = (2y)(1)+y(-1)+(-y)(1)=2y-y-y=0.

Also always satisfied.

Thus the only conditions are:

x=\pm \sqrt{\frac{3}{8}}, \qquad y=\pm \frac{1}{\sqrt2}.

Number of possible matrices:

$x$ has $2$ choices,
$y$ has $2$ choices.

Hence total number of matrices is

2\times 2 = 4.

Therefore the correct option is $\boxed{4}$ which is option B.

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