Animated Solution for Mathematics - Matrices and Determinants: If $A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$, then $\text{adj}(3A^2+12A)$ is equal to:

Why Brute Force is Your Enemy

Imagine you are sitting in the examination hall. You see a matrix $A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$ and a target expression: $\text{adj}(3A^2+12A)$.

Your first instinct might be to reach for your pen and start multiplying the matrix by itself. You think, 'I can do this, it's just a $2\times 2$ multiplication.'

Stop right there! This is the first trap of the JEE Advanced. While brute force will eventually yield the answer, it is a path paved with potential arithmetic errors and wasted time. In the world of competitive mathematics, we don't just solve problems; we dismantle them with elegance.

Instead of grinding through matrix multiplication, let us invoke one of the most beautiful theorems in linear algebra: the Cayley-Hamilton Theorem. This theorem tells us that every square matrix is a root of its own characteristic equation.

It essentially bridges the gap between the rigid world of matrices and the fluid world of polynomials. To find this equation, we look at the determinant of $A-\lambda I$. We are essentially asking: 'What values of $\lambda$ make the matrix $A-\lambda I$ singular?'

We set up the determinant:

Expanding this, we get $(2-\lambda )(1-\lambda )-(-3)(-4)=0$. Let's be careful with the signs here.

This simplifies to ${\lambda }^2-3\lambda +2-12=0$, which gives us the characteristic polynomial: ${\lambda }^2-3\lambda -10=0$. This is the 'DNA' of our matrix $A$.

By the Cayley-Hamilton theorem, we can replace $\lambda$ with $A$ and the constant term with the identity matrix $I$. Thus, $A^2-3A-10I=0$.

This is our golden key. It allows us to rewrite $A^2$ as $3A+10I$. We have successfully reduced a quadratic matrix term into a linear one!

Now, let's look at our target expression: $3A^2+12A$. Instead of calculating $A^2$, we substitute our new identity: $3(3A+10I)+12A$.

Look at how the complexity just melts away. Expanding this, we get $9A+30I+12A$.

Combining the like terms, we arrive at $21A+30I$. This is the power of the Cayley-Hamilton approach. We have bypassed the tedious multiplication entirely and arrived at a simple linear combination of $A$ and $I$.

Now, we calculate $21A$:

Adding $30I$, which is simply $\left[\begin{array}{cc}30& 0\\ 0& 30\end{array}\right]$, we get our final matrix $M$:

The Adjoint

We are at the finish line. The problem asks for the adjoint of $M$. For a $2\times 2$ matrix, we don't need to calculate cofactors and transposes manually.

We use the shortcut: swap the diagonal elements and negate the off-diagonal elements.

Taking our matrix $M=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$, we swap $72$ and $51$, and we flip the signs of $-63$ and $-84$. This gives us:

This matches Option 2 perfectly. Notice how we didn't just 'get the answer'; we navigated the problem using the structural properties of matrices.

This is the mindset of a topper. When you see a matrix polynomial, don't calculate—transform. Use the Cayley-Hamilton theorem to simplify, and always look for the elegant shortcut. Keep practicing, keep questioning, and keep falling in love with the logic behind the math!