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Step-by-Step Solution
Step 1: Express A in terms of the identity matrix and a nilpotent matrix
We are given
A =
\begin{pmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{pmatrix}.
We write
A = I + C,
where
I =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}, \quad
C =
\begin{pmatrix}
0 & -1 & 0 \\
0 & 0 & -1 \\
0 & 0 & 0
\end{pmatrix}.
Step 2: Observe the powers of the matrix C
Compute the successive powers of C:
C^2 =
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},
C^3 =
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} = 0.
Since C^3 = 0 , higher powers of C are also the zero matrix.
Step 3: Expand (I + C)^n
Using the binomial expansion and noting C^3=0 , any positive integer power of (I + C) can be written as
(I + C)^n = I + nC + \binom{n}{2} C^2.
Higher-order terms vanish because C^3=0 .
Step 4: Express A^{20} and A^7
Since A = I + C , we have:
A^{20} = (I + C)^{20} = I + 20C + \binom{20}{2} C^2.
Here, \binom{20}{2} = 190 .
A^7 = (I + C)^7 = I + 7C + \binom{7}{2} C^2.
Here, \binom{7}{2} = 21 .
Step 5: Form the matrix B and focus on the (1,3) entry
We are given
B = 7 A^{20} \;-\; 20 A^7 \;+\; 2I.
We want the entry b_{13} in the first row, third column. Note the following observations:
The matrix I has zero in the (1,3) position.
The matrix C has zero in the (1,3) position.
The matrix C^2 has 1 in the (1,3) position.
Thus, only the coefficients of C^2 from each term in B will contribute to b_{13} .
Step 6: Collect the coefficients of C^2
From
A^{20} = I + 20C + 190 C^2,
the coefficient of C^2 is 190 . From
A^7 = I + 7C + 21 C^2,
the coefficient of C^2 is 21 . Thus, the portion of B contributing to the (1,3) entry is
7 \bigl[\binom{20}{2}\bigr] C^2 - 20 \bigl[\binom{7}{2}\bigr] C^2.
Hence,
b_{13} = 7 \times 190 \;-\; 20 \times 21.
Calculate each term:
7 \times 190 = 1330,
20 \times 21 = 420.
Therefore,
b_{13} = 1330 \;-\; 420 = 910.
Final Answer
The value of b_{13} is 910 .
