Animated Solution for Mathematics - Matrices and Determinants: Let $x=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and $A=\left[\begin{array}{ccc}-1& 2& 3\\ 0& 1& 6\\ 0& 0& -1\end{array}\right]$. For $k\in \mathbb{N}$, if $X^T{A}^{k}X=33$, then $k$ is equal to:

Imagine standing before this matrix:

It looks like a standard $3\times 3$ matrix, but observe the lower triangle—it is filled with zeros. This is an upper triangular matrix, and in the context of JEE Advanced, this is a significant hint. It implies that the matrix is "well-behaved" under multiplication.

We are tasked to find $k$ such that $X^T{A}^{k}X=33$, where $X=[1,1,1{]}^T$. The term $A^k$ is the core of the problem. Do not be tempted to multiply $A$ by itself $k$ times; instead, we must identify a pattern.

Let us calculate $A^2$. When we multiply $A$ by itself, the diagonal elements become $1$, and the off-diagonal elements simplify significantly:

Notice how close this is to the identity matrix $I$. It is almost $I$, except for the $6$ in the top-right corner.

Now, let us calculate $A^4$ by multiplying $A^2$ with $A^2$. Because $A^2$ is so close to the identity, the multiplication is trivial. The diagonal remains $1$, and the top-right element becomes $12$.

The pattern is emerging: for $k=2$, the element is $6$ (which is $3\times 2$); for $k=4$, the element is $12$ (which is $3\times 4$).

We can now confidently generalize. For any even natural number $k$, the matrix $A^k$ takes the form:

This is the power of pattern recognition. We have reduced a complex matrix power to a simple algebraic expression involving $k$. This transformation is the "soul" of the problem.

Now, we return to the original equation: $X^T{A}^{k}X=33$. We know $X^T=[1,1,1]$ and $X=[1,1,1{]}^T$.

First, let us compute $A^{k}X$:

Finally, we multiply this result by $X^T$ on the left:

Setting this equal to $33$, we get $3k+3=33$, which simplifies to $3k=30$. Thus, the final result is: