Animated Solution for Mathematics - Matrices and Determinants: Let A be a matrix such that $A\cdot \left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ is a scalar matrix and $|3A|=108$. Then $A^2$ equals :

Welcome, fellow traveler on the JEE journey! Today, we are going to peel back the layers of a problem that looks like a daunting algebraic mess but is actually a beautiful exercise in the symmetry of matrix properties.

Imagine you are standing before a locked door. You have a key, but you don't know its shape. This problem is exactly like that—we are given a condition about a matrix $A$ and a known matrix $B$, and we must unlock the identity of $A^2$ without ever needing to solve for the individual elements of $A$ one by one.

The Hidden Geometry of the Scalar Matrix

We are told that $A\cdot B$ is a scalar matrix, where $B=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$. A scalar matrix is the simplest form of a square matrix—it is just a diagonal matrix where all diagonal elements are identical, like $kI$.

So, our starting point is the elegant equation $AB=kI$. This tells us that $A$ is essentially a scaled version of the inverse of $B$. If we can find $k^2$ and the inverse of $B$, the rest of the path clears up instantly.

The Power of Determinants

Now, let's look at the information provided: $|3A|=108$. Many students fall into the trap of thinking $|3A|=3|A|$.

But remember, when you multiply a matrix by a scalar $c$, you are multiplying every row by $c$. Since $A$ is a $2\times 2$ matrix, we have two rows, so $|3A|=3^2|A|=9|A|$. With this, we find that $|A|=\frac{108}{9}=12$.

Next, we look at the product $AB=kI$. By taking the determinant of both sides, we get $|AB|=|kI|$.

Using the multiplicative property of determinants, $|A|\cdot |B|=k^2|I|$. Since $|I|=1$, we have $k^2=|A|\cdot |B|$.

Calculating the determinant of $B$ is straightforward: $|B|=(1)(3)-(0)(2)=3$. Thus, $k^2=12\times 3=36$. We have successfully captured the scalar constant without ever needing to know $k$ itself!

The Final Transformation

We know $AB=kI$, which implies $A=k{B}^{-1}$. Squaring both sides gives us $A^2=k^2(B^{-1}{)}^2$.

We already have $k^2=36$, so we just need to compute $(B^{-1}{)}^2$. First, we find the inverse of $B$.

The adjoint of $B$ is $\left[\begin{array}{cc}3& -2\\ 0& 1\end{array}\right]$, and since $|B|=3$, we have $B^{-1}=\frac{1}{3}\left[\begin{array}{cc}3& -2\\ 0& 1\end{array}\right]$.

Squaring this matrix, we get:

Finally, we multiply this by our $k^2=36$:

Look at that result! By relying on the properties of determinants and the relationship between inverses and scalar matrices, we avoided a mountain of tedious algebra.

The JEE Advanced isn't just about calculation; it's about recognizing the structure of the problem. Keep this mindset, and you will find that even the most intimidating problems have a path of least resistance. You've got this!