Question

The total number of matrices
$A=\left( \begin{matrix}
   0 & 2x & 2x \\
   2y & y & -y \\
   1 & -1 & 1 \\
\end{matrix} \right),\left( x,y\in R,x\ne y \right)$ for which ${{A}^{T}}A=3{{I}_{3}}$ is:
(a)$6$
(b) 2
(c) 3
(d) 4

Answer 1

Hint: We have given a matrix A and we have to find the possible number of matrices which can satisfy this relation ${{A}^{T}}A=3{{I}_{3}}$. For that we have to find the transpose of matrix A which is the interchanging of rows with columns in a matrix then substitute the transpose of A and the matrix A in the given relation. And ${{I}_{3}}$ is a $3\times 3$ identity matrix in which all the elements in the main diagonal are 1 and the remaining elements are 0. Solve this equation and find the values of x and y and hence, find the possible number of matrices.

Complete step-by-step answer:
We have given a matrix A as follows:
$A=\left( \begin{matrix}
   0 & 2x & 2x \\
   2y & y & -y \\
   1 & -1 & 1 \\
\end{matrix} \right)$
The above matrix has come up with the restrictions on x and y as follows:
$\left( x,y\in R,x\ne y \right)$
And we have also given a condition and we have to use that condition to find the number of possible matrices and the condition is as follows:
${{A}^{T}}A=3{{I}_{3}}$……… Eq. (1)
In the above equation, ${{A}^{T}}$ is the transpose of matrix A and ${{I}_{3}}$ is the identity matrix of $3\times 3$ order which we have shown below:
${{A}^{T}}=\left( \begin{matrix}
   0 & 2y & 1 \\
   2x & y & -1 \\
   2x & -y & 1 \\
\end{matrix} \right)$
We have written the transpose of matrix A by interchanging columns of matrix A to rows.
${{I}_{3}}=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)$
Now, substituting the transpose of matrix A and the identity matrix in eq. (1) we get,
$\begin{align}
  & {{A}^{T}}A=3{{I}_{3}} \\
 & \Rightarrow \left( \begin{matrix}
   0 & 2y & 1 \\
   2x & y & -1 \\
   2x & -y & 1 \\
\end{matrix} \right)\left( \begin{matrix}
   0 & 2x & 2x \\
   2y & y & -y \\
   1 & -1 & 1 \\
\end{matrix} \right)=3\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
\end{align}$
Multiplying the left hand side of the two matrices and on the right hand side multiplying 3 with each element of the identity matrix we get,
$\begin{align}
  & \left( \begin{matrix}
   0\left( 0 \right)+\left( 2y \right)\left( 2y \right)+1\left( 1 \right) & 0\left( 2x \right)+2y\left( y \right)+1\left( -1 \right) & 0\left( 2x \right)+2y\left( -y \right)+1\left( 1 \right) \\
   2x\left( 0 \right)+y\left( 2y \right)+1\left( -1 \right) & 2x\left( 2x \right)+y\left( y \right)+\left( -1 \right)\left( -1 \right) & 2x\left( 2x \right)+y\left( -y \right)+1\left( -1 \right) \\
   2x\left( 0 \right)+2y\left( -y \right)+1\left( -1 \right) & 2x\left( 2x \right)+y\left( -y \right)+1\left( -1 \right) & 2x\left( 2x \right)+\left( -y \right)\left( -y \right)+1\left( 1 \right) \\
\end{matrix} \right)=\left( \begin{matrix}
   3 & 0 & 0 \\
   0 & 3 & 0 \\
   0 & 0 & 3 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   4{{y}^{2}}+1 & 2{{y}^{2}}-1 & -2{{y}^{2}}+1 \\
   2{{y}^{2}}-1 & 4{{x}^{2}}+{{y}^{2}}+1 & 4{{x}^{2}}-{{y}^{2}}-1 \\
   -2{{y}^{2}}-1 & 4{{x}^{2}}-{{y}^{2}}-1 & 4{{x}^{2}}+{{y}^{2}}+1 \\
\end{matrix} \right)=\left( \begin{matrix}
   3 & 0 & 0 \\
   0 & 3 & 0 \\
   0 & 0 & 3 \\
\end{matrix} \right) \\
\end{align}$
Now, when the two matrices are equal then each element of the matrix on the left hand side of the above equation is equal to the corresponding element written on the right hand side of the equation.
$\begin{align}
  & 4{{y}^{2}}+1=3, \\
 & 2{{y}^{2}}-1=0, \\
 & -2{{y}^{2}}+1=0, \\
 & 4{{x}^{2}}+{{y}^{2}}+1=3, \\
 & 4{{x}^{2}}-{{y}^{2}}-1=0 \\
 & -2{{y}^{2}}-1=0, \\
\end{align}$
Finding the solutions of the above equation we get,
Solving the first equation in the equations written above we get,
$\begin{align}
  & 4{{y}^{2}}+1=3 \\
 & \Rightarrow 4{{y}^{2}}=2 \\
 & \Rightarrow {{y}^{2}}=\dfrac{2}{4}=\dfrac{1}{2} \\
\end{align}$
Solving the second equation in the equations written above we get,
$\begin{align}
  & 2{{y}^{2}}-1=0 \\
 & \Rightarrow {{y}^{2}}=\dfrac{1}{2} \\
\end{align}$
Solving the third equation in the equations written above we get,
$\begin{align}
  & -2{{y}^{2}}+1=0 \\
 & \Rightarrow 2{{y}^{2}}=1 \\
 & \Rightarrow {{y}^{2}}=\dfrac{1}{2} \\
\end{align}$
As you can see that we are getting the same value of y in all the three equations so the value of y is:
$\begin{align}
  & {{y}^{2}}=\dfrac{1}{2} \\
 & \Rightarrow y=\pm \dfrac{1}{\sqrt{2}} \\
\end{align}$
Substituting the value of ${{y}^{2}}$ in the fourth equation given above we get,
$\begin{align}
  & 4{{x}^{2}}+\left( \dfrac{1}{2} \right)+1=3 \\
 & \Rightarrow 4{{x}^{2}}+\dfrac{3}{2}=3 \\
 & \Rightarrow 4{{x}^{2}}=3-\dfrac{3}{2} \\
 & \Rightarrow 4{{x}^{2}}=\dfrac{3}{2} \\
\end{align}$
Dividing 4 on both the sides we get,
${{x}^{2}}=\dfrac{3}{8}$
Taking square root on both the sides we get,
$\begin{align}
  & x=\pm \sqrt{\dfrac{3}{8}} \\
 & \Rightarrow x=\pm \dfrac{1}{2}\sqrt{\dfrac{3}{2}} \\
\end{align}$
From the above equations, we have got two values of x and two values of y which is shown below:\[x=\pm \dfrac{1}{2}\sqrt{\dfrac{3}{2}},y=\pm \dfrac{1}{\sqrt{2}}\]
As you can see that x and y values are not equal to each other so it is satisfying the condition given in the above problem.
Now, we have to find the possible set of x and y which can be put in the matrix A. The possible sets are:
$\begin{align}
  & \left( x,y \right) \\
 & =\left( \dfrac{1}{2}\sqrt{\dfrac{3}{2}},\dfrac{1}{\sqrt{2}} \right),\left( -\dfrac{1}{2}\sqrt{\dfrac{3}{2}},\dfrac{1}{\sqrt{2}} \right),\left( \dfrac{1}{2}\sqrt{\dfrac{3}{2}},-\dfrac{1}{\sqrt{2}} \right),\left( -\dfrac{1}{2}\sqrt{\dfrac{3}{2}},-\dfrac{1}{\sqrt{2}} \right) \\
\end{align}$
As you can see that 4 sets of possibilities of $\left( x,y \right)$ pair are possible that could be substituted in the matrix A so 4 matrices are possible.
Hence, the correct option is (d).

Note: The most common mistake that could happen is the calculation mistake because as you can see rigorous calculation is involved in solving the multiplication of two matrices along with that while multiplying 3 to the identity matrix you could have mistakenly multiply just any row or column because you might have mix the concepts of multiplication of a constant to a determinant and a matrix. The concept is in determinants, if we multiply any constant to the determinant then only one of the row or column gets multiplied whereas in the matrices when we multiply a constant to it then we have to multiply all the elements written in the matrix. The mistake is you might think that while multiplying a constant to a matrix we just have to multiply any row or column which is actually the concept for determinants.