Key Concepts:

This problem requires us to work with matrix powers and matrix multiplication. The central idea is to:

1. Identify a pattern for the $k$-th power of matrix $A$ ($A^k$). This often involves calculating the first few powers ($A^2, A^3, \dots$) and looking for a discernible structure, especially considering the parity of $k$.
2. Perform matrix multiplication for the expression $X'A^kX$. $X'$ denotes the transpose of matrix $X$.
3. Solve the resulting equation for $k$ using the given condition.

Step 1: Understand the Given Matrices and the Expression

We are given:

• Matrix $X$: A column vector of ones. $X=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$
• Its transpose, $X'$: A row vector of ones. $X'=\left[\begin{array}{lll} 1 & 1 & 1 \end{array}\right]$
• Matrix $A$: An upper triangular matrix. $A=\left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$
• The condition: $X' A^k X = 33$, where $k \in \mathbb{N}$ (natural numbers, i.e., positive integers).

Our goal is to find the value of $k$.

Step 2: Calculate Initial Powers of Matrix A to Identify a Pattern

To find a general expression for $A^k$, we compute $A^2, A^3$, and observe their structure.

• Calculate $A^2 = A \cdot A$: $A^2 = \left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right] \left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$ Performing the matrix multiplication:

  • $(A^2)_{11} = (-1)(-1) + 2(0) + 3(0) = 1$
  • $(A^2)_{12} = (-1)(2) + 2(1) + 3(0) = -2 + 2 + 0 = 0$
  • $(A^2)_{13} = (-1)(3) + 2(6) + 3(-1) = -3 + 12 - 3 = 6$
  • $(A^2)_{21} = 0(-1) + 1(0) + 6(0) = 0$
  • $(A^2)_{22} = 0(2) + 1(1) + 6(0) = 1$
  • $(A^2)_{23} = 0(3) + 1(6) + 6(-1) = 6 - 6 = 0$
  • $(A^2)_{31} = 0(-1) + 0(0) + (-1)(0) = 0$
  • $(A^2)_{32} = 0(2) + 0(1) + (-1)(0) = 0$
  • $(A^2)_{33} = 0(3) + 0(6) + (-1)(-1) = 1$ So, $A^2 = \left[\begin{array}{ccc} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ Observation: $A^2$ is an identity matrix with a $6$ in the $(1,3)$ position. This is a special structure. Let $I$ be the identity matrix and $J = \left[\begin{array}{ccc} 0 & 0 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$. Then $A^2 = I + J$. Notice that $J^2 = \left[\begin{array}{ccc} 0 & 0 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \left[\begin{array}{ccc} 0 & 0 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = 0$. This means $J$ is a nilpotent matrix of order 2.

• Calculate $A^3 = A^2 \cdot A$: $A^3 = \left[\begin{array}{ccc} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$ Performing the matrix multiplication:

  • $(A^3)_{11} = 1(-1) + 0(0) + 6(0) = -1$
  • $(A^3)_{12} = 1(2) + 0(1) + 6(0) = 2$
  • $(A^3)_{13} = 1(3) + 0(6) + 6(-1) = 3 - 6 = -3$
  • $(A^3)_{21} = 0(-1) + 1(0) + 0(0) = 0$
  • $(A^3)_{22} = 0(2) + 1(1) + 0(0) = 1$
  • $(A^3)_{23} = 0(3) + 1(6) + 0(-1) = 6$
  • $(A^3)_{31} = 0(-1) + 0(0) + 1(0) = 0$
  • $(A^3)_{32} = 0(2) + 0(1) + 1(0) = 0$
  • $(A^3)_{33} = 0(3) + 0(6) + 1(-1) = -1$ So, $A^3 = \left[\begin{array}{ccc} -1 & 2 & -3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$ Observation: $A^3$ is very similar to $A$, but the $(1,3)$ element has changed sign and value.

Step 3: Deduce a General Form for $A^k$

We observe that the pattern for $A^k$ depends on whether $k$ is even or odd.

• Case 1: $k$ is an even number. Let $k = 2m$ for some natural number $m$. We have $A^2 = I + J$, where $J^2 = 0$. Then, $A^k = A^{2m} = (A^2)^m = (I+J)^m$. Using the binomial expansion (since $I$ and $J$ commute, $IJ = JI = J$): $(I+J)^m = I^m + \binom{m}{1}I^{m-1}J + \binom{m}{2}I^{m-2}J^2 + \dots + J^m$ Since $J^2 = 0$, all terms with $J^n$ for $n \ge 2$ are zero. Thus, $(I+J)^m = I + mJ$. Substituting $I$ and $J$: $A^k = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + m \left[\begin{array}{ccc} 0 & 0 & 6 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 6m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ Since $k=2m$, we have $m = k/2$. So, for even $k$: $A^k = \left[\begin{array}{ccc} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

• Case 2: $k$ is an odd number. Let $k = 2m+1$ for some non-negative integer $m$. We can write $A^k = A^{2m+1} = A^{2m} \cdot A$. Using the result for $A^{2m}$ from Case 1 ($A^{2m} = \left[\begin{array}{ccc} 1 & 0 & 6m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$): $A^k = \left[\begin{array}{ccc} 1 & 0 & 6m \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$ Performing the matrix multiplication:

  • $(A^k)_{11} = 1(-1) + 0(0) + 6m(0) = -1$
  • $(A^k)_{12} = 1(2) + 0(1) + 6m(0) = 2$
  • $(A^k)_{13} = 1(3) + 0(6) + 6m(-1) = 3 - 6m$
  • $(A^k)_{21} = 0$
  • $(A^k)_{22} = 1$
  • $(A^k)_{23} = 6$
  • $(A^k)_{31} = 0$
  • $(A^k)_{32} = 0$
  • $(A^k)_{33} = -1$ So, for odd $k$: $A^k = \left[\begin{array}{ccc} -1 & 2 & 3 - 6m \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$ Since $k = 2m+1$, we have $2m = k-1$. Substituting $6m = 3(2m) = 3(k-1)$: $A^k = \left[\begin{array}{ccc} -1 & 2 & 3 - 3(k-1) \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right] = \left[\begin{array}{ccc} -1 & 2 & 3 - 3k + 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right] = \left[\begin{array}{ccc} -1 & 2 & 6 - 3k \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$

Step 4: Evaluate $X'A^kX$ for each case and solve for $k$

Now we use the derived forms of $A^k$ to calculate $X'A^kX$ and equate it to 33.

• Case 1: $k$ is even. Substitute $A^k = \left[\begin{array}{ccc} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$ into $X'A^kX$: $X'A^kX = \left[\begin{array}{lll} 1 & 1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$ First, calculate $X'A^k$: $X'A^k = \left[\begin{array}{lll} 1 \cdot 1 + 1 \cdot 0 + 1 \cdot 0 & 1 \cdot 0 + 1 \cdot 1 + 1 \cdot 0 & 1 \cdot (3k) + 1 \cdot 0 + 1 \cdot 1 \end{array}\right]$ $X'A^k = \left[\begin{array}{lll} 1 & 1 & 3k+1 \end{array}\right]$ Now, calculate $(X'A^k)X$: $X'A^kX = \left[\begin{array}{lll} 1 & 1 & 3k+1 \end{array}\right] \left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] = 1(1) + 1(1) + (3k+1)(1) = 1 + 1 + 3k + 1 = 3k+3$ Equate this to 33: $3k+3 = 33$ $3k = 30$ $k = 10$ Since $k=10$ is an even natural number, this is a valid solution.

• Case 2: $k$ is odd. Substitute $A^k = \left[\begin{array}{ccc} -1 & 2 & 6 - 3k \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right]$ into $X'A^kX$: $X'A^kX = \left[\begin{array}{lll} 1 & 1 & 1 \end{array}\right] \left[\begin{array}{ccc} -1 & 2 & 6 - 3k \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{array}\right] \left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$ First, calculate $X'A^k$: $X'A^k = \left[\begin{array}{lll} 1(-1)+1(0)+1(0) & 1(2)+1(1)+1(0) & 1(6-3k)+1(6)+1(-1) \end{array}\right]$ $X'A^k = \left[\begin{array}{lll} -1 & 3 & 6-3k+6-1 \end{array}\right] = \left[\begin{array}{lll} -1 & 3 & 11-3k \end{array}\right]$ Now, calculate $(X'A^k)X$: $X'A^kX = \left[\begin{array}{lll} -1 & 3 & 11-3k \end{array}\right] \left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] = -1(1) + 3(1) + (11-3k)(1) = -1 + 3 + 11 - 3k = 13 - 3k$ Equate this to 33: $13 - 3k = 33$ $-3k = 20$ $k = -\frac{20}{3}$ Since $k$ must be a natural number ($k \in \mathbb{N}$), this solution is not valid.

Step 5: Conclude the Value of k

From our analysis, the only valid value for $k$ is $10$.

Tips and Common Mistakes:

• Careful Matrix Multiplication: Matrix multiplication is prone to small errors. Double-check each element calculation.
• Pattern Recognition: When finding $A^k$, compute enough initial powers ($A^2, A^3, A^4$) to confidently establish a pattern. Look for differences based on the parity of $k$.
• Special Matrix Forms: Recognize if a matrix can be expressed as $I+N$ where $N$ is nilpotent (i.e., $N^p=0$ for some $p$). This can greatly simplify calculating powers using the binomial theorem.
• Check Conditions: Always ensure your final answer for $k$ satisfies all given conditions, such as $k \in \mathbb{N}$ and the parity assumed for the pattern.

Summary:

We systematically determined the general form of $A^k$ by analyzing $A^2$ and $A^3$, recognizing different patterns for even and odd powers of $k$. We then used these forms to calculate $X'A^kX$ and solved for $k$ by equating it to 33. The only natural number solution that satisfies the condition is $k=10$.

The final answer is $\boxed{10}$.