1. Key Concepts and Formulas

• Cayley-Hamilton Theorem: Every square matrix satisfies its own characteristic equation. For a $3 \times 3$ matrix $A$, if its characteristic polynomial is $P(\lambda) = \det(A - \lambda I)$, then $P(A) = 0$.
• Matrix Recurrence Relations: If a matrix power $A^n$ is defined by a linear recurrence relation involving lower powers of $A$, such as $A^n = c_1 A^{n-1} + c_2 A^{n-2} + \dots + c_k A^{n-k} + C$, where $C$ is a constant matrix, we can often find a general formula for $A^n$ by identifying patterns or using techniques similar to solving linear scalar recurrence relations.
• Sum of Elements of a Matrix: For matrices $M$ and $N$ of the same dimensions, and a scalar $c$, the sum of elements $S(M+N) = S(M) + S(N)$ and $S(cM) = cS(M)$.

2. Step-by-Step Solution

Step 1: Calculate the square of matrix A ($A^2$). To use the given recurrence relation and later compute $A^{50}$, we first need to calculate $A^2$. Given $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$. $A^2 = A \times A = \left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right] \left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$ $A^2 = \left[\begin{array}{ccc} (1)(1)+(0)(1)+(0)(0) & (1)(0)+(0)(0)+(0)(1) & (1)(0)+(0)(1)+(0)(0) \\ (1)(1)+(0)(1)+(1)(0) & (1)(0)+(0)(0)+(1)(1) & (1)(0)+(0)(1)+(1)(0) \\ (0)(1)+(1)(1)+(0)(0) & (0)(0)+(1)(0)+(0)(1) & (0)(0)+(1)(1)+(0)(0) \end{array}\right]$ $A^2 = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right]$

Step 2: Verify the consistency of the given recurrence relation with the Cayley-Hamilton Theorem. The given recurrence is $A^n = A^{n-2} + A^2 - I$ for $n \geqslant 3$. Let's find the characteristic equation of $A$. $P(\lambda) = \det(A - \lambda I) = \left|\begin{array}{ccc}1-\lambda & 0 & 0 \\ 1 & -\lambda & 1 \\ 0 & 1 & -\lambda\end{array}\right|$ Expanding along the first row: $P(\lambda) = (1-\lambda)((-\lambda)(-\lambda) - (1)(1)) - 0 + 0 = (1-\lambda)(\lambda^2 - 1)$ $P(\lambda) = (1-\lambda)(\lambda-1)(\lambda+1) = -(\lambda-1)^2(\lambda+1)$ By the Cayley-Hamilton Theorem, $A$ satisfies its characteristic equation $P(A) = 0$: $-(A-I)^2(A+I) = 0$ $(A^2 - 2A + I)(A+I) = 0$ $A^3 + A^2 - 2A^2 - 2A + A + I = 0$ $A^3 - A^2 - A + I = 0$ This implies $A^3 = A^2 + A - I$.

Now, let's check the given recurrence for $n=3$: $A^3 = A^{3-2} + A^2 - I = A + A^2 - I$. This matches the result from the Cayley-Hamilton Theorem, confirming the consistency of the given recurrence relation with the matrix $A$.

Step 3: Find a general expression for $A^n$ for even $n$. The recurrence relation is $A^n = A^{n-2} + A^2 - I$ for $n \ge 3$. Let $C = A^2 - I$. This is a constant matrix. $C = \left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] - \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{array}\right]$. The recurrence becomes $A^n = A^{n-2} + C$.

We need to find $A^{50}$, which is an even power. Let $n=2k$. For $n=4$ (i.e., $k=2$): $A^4 = A^{4-2} + C = A^2 + C$. For $n=6$ (i.e., $k=3$): $A^6 = A^{6-2} + C = A^4 + C = (A^2 + C) + C = A^2 + 2C$. Continuing this pattern, for any even $n=2k$ where $k \ge 2$: $A^{2k} = A^2 + (k-1)C$.

We need to find $A^{50}$. Here $n=50$, so $2k=50 \implies k=25$. Since $k=25 \ge 2$, the formula applies: $A^{50} = A^2 + (25-1)C = A^2 + 24C$.

Step 4: Substitute and calculate $A^{50}$. Substitute the values of $A^2$ and $C$ into the expression for $A^{50}$: $A^{50} = \left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] + 24 \left[\begin{array}{ccc}0 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{array}\right]$ $A^{50} = \left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] + \left[\begin{array}{ccc}24 \cdot 0 & 24 \cdot 0 & 24 \cdot 0 \\ 24 \cdot 1 & 24 \cdot 0 & 24 \cdot 0 \\ 24 \cdot 1 & 24 \cdot 0 & 24 \cdot 0\end{array}\right]$ $A^{50} = \left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] + \left[\begin{array}{ccc}0 & 0 & 0 \\ 24 & 0 & 0 \\ 24 & 0 & 0\end{array}\right]$ $A^{50} = \left[\begin{array}{ccc}1+0 & 0+0 & 0+0 \\ 1+24 & 1+0 & 0+0 \\ 1+24 & 0+0 & 1+0\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 25 & 1 & 0 \\ 25 & 0 & 1\end{array}\right]$

Step 5: Calculate the sum of all elements of $A^{50}$. The sum of all elements of $A^{50}$ is the sum of all entries in the matrix: Sum $= 1 + 0 + 0 + 25 + 1 + 0 + 25 + 0 + 1 = 1 + 25 + 1 + 25 + 1 = 53$.

Common Mistakes & Tips

• Careful Matrix Multiplication: Matrix multiplication is prone to errors. Double-check each element calculation, especially for off-diagonal terms.
• Correct Base Case for Recurrence: Ensure the derived general formula for $A^n$ is valid for the specific $n$ value (e.g., $n=50$ in this case) and that the base case of the recurrence is correctly applied.
• Distinguishing between $A^2$ and $A^2-I$: The constant matrix $C$ in the recurrence is $A^2-I$, not just $A^2$. Misinterpreting this can lead to incorrect patterns.

Summary

We first calculated $A^2$ and then used the Cayley-Hamilton theorem to confirm the given recurrence relation $A^n = A^{n-2} + A^2 - I$ for $n=3$. By setting $C = A^2 - I$, the recurrence simplified to $A^n = A^{n-2} + C$. For even powers, this formed an arithmetic progression $A^{2k} = A^2 + (k-1)C$. Substituting $k=25$ for $A^{50}$, we found $A^{50} = A^2 + 24C$. Finally, we computed the elements of $A^{50}$ and summed them, arriving at 53.

The final answer is $\boxed{\text{53}}$