Question

Let
 \(x = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\) and \(A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}\)
For k ∈ N, if X’A^kX = 33, then k is equal to ____ .

Answer 1

Correct Answer: 10

To find \(k\) such that \(X’A^{k}X = 33\), we first calculate \(A^{k}\) and then evaluate the matrix operation.

Given:
\(\)
\(A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}\) 

Step 1: Calculate \(A^{k}\) using diagonalization since \(A\) is upper triangular.

The eigenvalues of \(A\) (diagonal elements) are \(-1, 1, -1\).

Step 2: Find \(X’AX\)

\(X' = \begin{bmatrix}1 & 1 & 1\end{bmatrix}\)
So, \(AX = \begin{bmatrix}-1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1\end{bmatrix} \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}4 \\ 7 \\ -1\end{bmatrix}\)

Then, calculate \(X'AX = \begin{bmatrix}1 & 1 & 1\end{bmatrix} \begin{bmatrix}4 \\ 7 \\ -1\end{bmatrix} = 10\)

Step 3: Evaluate \(X'A^{k}X\) iteratively until it equals 33 with \(A^{k}X\).

Since \(X'AX = 10\), we need \(10^{k} = 33\). Trying integers:

• If \(k=10\), calculate \(10^{10}\) which well exceeds 33; re-examine steps or calculations.

 

Step 4: Reassess multiple calculations considering matrix properties and overlaps in upper triangular matrices.

Retesting and reconfiguring involves direct matrix powers:\(A^{2} =\begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & 1\end{bmatrix}\), still trials until:

-Finding errors usually arise repeating sequence values; in rigorous processing adjust output calculations \(k=3\) as powers indicate simplified results generated cause repeat.

Confirm:\(k=10\) outcomes don't apply here.

Answer 2

Given that, 
\(A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}\)
\(A^2 = \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
\(A^4 = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}\)
\(A^k = \begin{bmatrix} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
So, \(X^′A^kX= \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\)
\(⇒X^′A^kX=[3k+3]\)
⇒ [3k + 3] = 33 (here it shall be [33] as matrix can’t be equal to a scalar)
i.e. [3k + 3] = 33
3k + 3 = [33] ⇒ k = 10
If k is odd and apply above process, we don’t get odd value of k
∴ k = 10
So, the answer is 10.