Animated Solution for Mathematics - Matrices and Determinants: The total number of matrices $A=\left(\begin{array}{ccc}0& 2y& 1\\ 2x& y& -1\\ 2x& -y& 1\end{array}\right),(x,y\in R,x\ne y)$ for which $A^{T}A=3I_3$ is :-

Welcome, future engineer! Today, we are going to unravel the mystery of a matrix equation that, at first glance, might look like a daunting algebraic mess.

We are given a matrix $A=\left(\begin{array}{ccc}0& 2y& 1\\ 2x& y& -1\\ 2x& -y& 1\end{array}\right)$ and the condition $A^{T}A=3I_3$.

This condition is not just a random equation; it is a powerful geometric statement. It tells us that the columns of $A$ are orthogonal to each other and each has a norm of $\sqrt{3}$.

The Transpose and the Product

To use our condition, we first need to find the transpose of matrix $A$. We simply interchange the rows and columns, resulting in:

Now, we compute the product $A^{T}A$. We take the dot product of the rows of the first matrix with the columns of the second matrix to obtain a new $3\times 3$ matrix.

The Magic of Cancellation

Let's compute the diagonal elements. For the $(1,1)$ element, we multiply the first row of $A^T$ with the first column of $A$:

Moving to the middle diagonal element, the $(2,2)$ element, we multiply the second row of $A^T$ with the second column of $A$:

Finally, for the $(3,3)$ element, we multiply the third row of $A^T$ with the third column of $A$:

If you calculate the off-diagonal elements, you will see they all beautifully cancel out to zero. We are left with the following diagonal matrix:

The Final Hurdle

Now, we equate our resulting matrix to $3I_3=\left(\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right)$. By comparing the corresponding elements, we obtain two simple equations:

Taking the square root, we get two possible values for $x$, namely $x=\pm \sqrt{\frac{3}{8}}$, and two possible values for $y$, namely $y=\pm \sqrt{\frac{1}{2}}$.

Since $\frac{3}{8} eq \frac{1}{2}$, the condition $x eq y$ is always satisfied for any combination of these values.

We have two independent choices for $x$ (positive or negative) and two independent choices for $y$ (positive or negative).

Since each combination of $x$ and $y$ gives a unique matrix $A$, the total number of possible matrices is:

You have successfully navigated the trap and found the solution. Keep this mindset for your JEE Advanced journey—always look for the underlying structure, and the math will follow!