Animated Solution for Mathematics - Matrices and Determinants: Let the matrix $A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$ and the matrix $B_0=A^{49}+2A^{98}$. If $B_n=\text{adj}(B_{n-1})$ for all $n\ge 1$, then $\det (B_4)$ is equal to :

A Journey into Matrix Recursion

When you first encounter a problem like this, it is natural to feel a surge of anxiety. You see $A^{49}$ and $A^{98}$, and your mind immediately jumps to the nightmare of multiplying a matrix by itself forty-nine times.

But stop. Take a breath. In the world of JEE Advanced, a problem that looks like a brute-force calculation is almost always a trap. There is a hidden elegance, a rhythmic pattern waiting to be uncovered. Let us embark on this journey together.

Decoding the Permutation Matrix

Look at the matrix $A=\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$. It is sparse, clean, and deliberate. This is a permutation matrix.

If you visualize the standard basis vectors $e_1,e_2,e_3$ in ${\mathbb{R}}^3$, this matrix acts as a cyclic operator. It maps $e_1\to e_3$, $e_3\to e_2$, and $e_2\to e_1$.

Because it is a cycle of three, we suspect that $A^3$ must return us to the identity matrix $I$. Let us verify:

This is our breakthrough! The matrix is periodic with a period of $3$. This means any power of $A$ can be reduced using simple division.

For $A^{49}$, we write $49=3\times 16+1$, so $A^{49}=(A^3{)}^{16}\cdot A=I^{16}\cdot A=A$. Similarly, $98=3\times 32+2$, so $A^{98}=A^2$. The terrifying exponents have vanished, leaving us with $B_0=A+2A^2$.

The Determinant of $B_0$

Now that we have $B_0=A+2A^2$, we construct the matrix explicitly:

Calculating the determinant of this $3\times 3$ matrix is straightforward. Expanding along the first row:

We see the number $9$, which is $3^2$. Keep this value close; it is the seed of our final answer.

The Adjoint Recursion

Here is where the problem shifts from algebra to a beautiful recursive sequence. We are given $B_n=\text{adj}(B_{n-1})$. We need $\det (B_4)$.

Recall the golden property of the adjoint matrix for an $n\times n$ matrix $M$:

Since our matrix is $3\times 3$, the exponent is $3-1=2$. Thus, $\det (\text{adj}(M))=(\det M{)}^2$.

This is the engine of our solution. Every time we move from $B_{n-1}$ to $B_n$, we are squaring the determinant of the previous matrix. Let us trace the chain reaction:

- $\det (B_1)=(\det B_0{)}^2$ - $\det (B_2)=(\det B_1{)}^2=((\det B_0{)}^2{)}^2=(\det B_0{)}^4$ - $\det (B_3)=(\det B_2{)}^2=((\det B_0{)}^4{)}^2=(\det B_0{)}^8$ - $\det (B_4)=(\det B_3{)}^2=((\det B_0{)}^8{)}^2=(\det B_0{)}^{16}$

The Final Calculation

We have arrived at the finish line. We know $\det (B_0)=3^2$. Substituting this into our general formula for $\det (B_4)$:

Look at that result. It is clean, precise, and derived not through brute force, but through the elegant application of matrix properties. You didn't just solve a problem; you decoded a system. Remember this feeling—the moment the complexity collapses into simplicity—because that is the true heart of mathematics. The final answer is $3^{32}$.