Key Concept: Binomial Theorem for Matrices and Nilpotent Matrices

This problem involves calculating high powers of a matrix. Directly multiplying matrices 20 or 7 times is computationally intensive and error-prone. The key insight is to recognize the special structure of matrix $A$ and express it as a sum of the identity matrix $I$ and a nilpotent matrix $N$.

The Binomial Theorem for Matrices states that if two matrices $X$ and $Y$ commute (i.e., $XY = YX$), then for any positive integer $n$: $(X+Y)^n = \sum_{k=0}^n \binom{n}{k} X^{n-k} Y^k = \binom{n}{0}X^n Y^0 + \binom{n}{1}X^{n-1}Y^1 + \binom{n}{2}X^{n-2}Y^2 + \dots + \binom{n}{n}X^0 Y^n$ A nilpotent matrix $N$ is a square matrix such that $N^k = \mathbf{0}$ (the zero matrix) for some positive integer $k$. The smallest such $k$ is called the index of nilpotency. If $N$ is nilpotent, many terms in its polynomial expansion (like in the Binomial Theorem) will vanish, significantly simplifying calculations.

Step 1: Decompose Matrix A into I + N

First, let's analyze the given matrix $A$: A = \left( {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr 0 & 0 & 1 \cr } } \right) Notice that $A$ is an upper triangular matrix with all diagonal elements equal to 1. Such matrices can often be expressed as the sum of the identity matrix $I$ and a strictly upper triangular (and thus nilpotent) matrix $N$. Let $A = I + N$. Then $N = A - I$: N = \left( {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr 0 & 0 & 1 \cr } } \right) - \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right) = \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right)

Step 2: Determine the Nilpotency Index of N

Now, let's calculate powers of $N$ to find its nilpotency index. N^2 = N \cdot N = \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right) \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right) = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) N^3 = N^2 \cdot N = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) = \mathbf{0} Since $N^3 = \mathbf{0}$, $N$ is a nilpotent matrix of index 3. This means $N^k = \mathbf{0}$ for all $k \ge 3$. This property is crucial for simplifying higher powers of $A$.

Step 3: Calculate Powers of A using the Binomial Theorem

Since $A = I + N$ and the identity matrix $I$ commutes with any matrix $N$ (i.e., $IN = NI = N$), we can apply the Binomial Theorem directly: $A^n = (I+N)^n = \binom{n}{0}I^n N^0 + \binom{n}{1}I^{n-1}N^1 + \binom{n}{2}I^{n-2}N^2 + \binom{n}{3}I^{n-3}N^3 + \dots$ Because $N^3 = \mathbf{0}$, all terms with $N^k$ for $k \ge 3$ will be zero. Thus, for any $n \ge 2$: $A^n = \binom{n}{0}I + \binom{n}{1}N + \binom{n}{2}N^2$ $A^n = I + nN + \frac{n(n-1)}{2}N^2$

Now, let's calculate $A^{20}$ and $A^7$: For $A^{20}$ (here $n=20$): $A^{20} = I + 20N + \frac{20(20-1)}{2}N^2 = I + 20N + \frac{20 \times 19}{2}N^2 = I + 20N + 190N^2$ For $A^7$ (here $n=7$): $A^7 = I + 7N + \frac{7(7-1)}{2}N^2 = I + 7N + \frac{7 \times 6}{2}N^2 = I + 7N + 21N^2$

Step 4: Substitute into the Expression for B

The matrix $B$ is given by $B = 7A^{20} - 20A^7 + 2I$. Substitute the expressions for $A^{20}$ and $A^7$: $B = 7(I + 20N + 190N^2) - 20(I + 7N + 21N^2) + 2I$

Step 5: Simplify Matrix B

Distribute the scalar coefficients and group terms by $I$, $N$, and $N^2$: $B = (7I + 140N + 1330N^2) - (20I + 140N + 420N^2) + 2I$ $B = (7 - 20 + 2)I + (140 - 140)N + (1330 - 420)N^2$ $B = (-11)I + (0)N + (910)N^2$ So, $B = -11I + 910N^2$.

Step 6: Find the Element $b_{13}$

We need to find $b_{13}$, which is the element in the first row and third column of matrix $B$. Let's look at the $(1,3)$ elements of $I$ and $N^2$: I = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right) \implies (I)_{13} = 0 N^2 = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) \implies (N^2)_{13} = 1 Now, substitute these values into the expression for $B$: $b_{13} = -11(I)_{13} + 910(N^2)_{13}$ $b_{13} = -11(0) + 910(1)$ $b_{13} = 0 + 910$ $b_{13} = 910$

Common Mistakes to Avoid:

• Incorrectly calculating powers of N: Double-check matrix multiplication, especially for off-diagonal elements.
• Forgetting to distribute scalar multiples: Ensure all terms inside the parentheses are multiplied by their respective coefficients.
• Assuming nilpotency for non-strictly triangular matrices: Only strictly upper or strictly lower triangular matrices are guaranteed to be nilpotent.
• Applying Binomial Theorem without checking commutativity: The theorem $(X+Y)^n$ only works if $XY = YX$. Here, $I$ commutes with any matrix $N$, so it's valid.

Summary and Key Takeaway: The final answer is $\boxed{910}$. This problem demonstrates the power of recognizing special matrix structures. By decomposing $A$ into $I+N$ where $N$ is nilpotent, we could use the Binomial Theorem to efficiently calculate high powers of $A$. This method avoids tedious matrix multiplications and simplifies the entire problem into a polynomial evaluation for $I$, $N$, and $N^2$. This approach is particularly effective for matrices that are "perturbations" of the identity matrix.