Let $\mathrm{X}=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$ and $\mathrm{A}=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{array}\right]$. For $\mathrm{k} \in \mathbb{N}$, if $\mathrm{X}^{\prime} \mathrm{A}^{k} \mathrm{X}=33$, then k is equal to:

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Matrices

Let

X = 111

and

A = - 1 00 210 36 - 1

. For

k \in N

, if

X^{'} A^{k} X = 33

, then k is equal to:

Answer: 10

Squaring gives

A^{2} = 100010601

, and continuing to square shows that for even

k

A^{k} = 100010 3 k 01

. With

X = [1, 1, 1]^{T}

, the scalar

X^{T} A^{k} X

equals the sum of all entries of

A^{k}

, namely

3 + 3 k

. Setting

3 + 3 k = 33

gives

k = 10

, which is even and therefore consistent with the form used.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer