Question

Consider the following matrices A and B.
\[ A = \begin{pmatrix} 1 & 2 & 0 & 0 \\ 3 & 4 & 0 & 0 \\ 0 & 5 & 5 & 0 \\ 0 & 0 & 6 & 7 \\ 0 & 0 & 8 & 9 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 10 & 11 & 0 & 0 & 0 \\ 12 & 13 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 15 & 16 & 0 \\ 0 & 0 & 17 & 18 & 0 \end{pmatrix} \]

If \( C = AB \), the sum of the diagonal elements of \( C \) is ..............

Hint:

To find the sum of the diagonal elements of a matrix product, multiply corresponding rows and columns and sum the diagonal elements of the resultant matrix.

Answer 1

The matrix multiplication \( C = AB \) involves multiplying the rows of matrix \( A \) with the columns of matrix \( B \). We need to find the sum of the diagonal elements of the resultant matrix \( C \).
To calculate the diagonal elements of \( C \), we perform the following steps:
Step 1: Multiply matrices \( A \) and \( B \).
Matrix \( A \) is \( 5 \times 4 \), and matrix \( B \) is \( 4 \times 5 \), so the resultant matrix \( C \) will be a \( 5 \times 5 \) matrix.
To find the diagonal elements, we compute:
- \( C_{11} = (1 \times 10) + (2 \times 12) + (0 \times 0) + (0 \times 0) = 10 + 24 = 34 \)
- \( C_{22} = (3 \times 11) + (4 \times 13) + (0 \times 0) + (0 \times 0) = 33 + 52 = 85 \)
- \( C_{33} = (0 \times 10) + (5 \times 12) + (5 \times 4) + (0 \times 0) = 60 + 20 = 80 \)
- \( C_{44} = (0 \times 10) + (0 \times 12) + (6 \times 4) + (7 \times 15) = 24 + 105 = 129 \)
- \( C_{55} = (0 \times 10) + (0 \times 12) + (8 \times 4) + (9 \times 15) = 32 + 135 = 167 \)
Step 2: Sum the diagonal elements of \( C \).
Now we sum the diagonal elements: \[ 34 + 85 + 80 + 129 + 167 = 495 \] Final Answer: \[ \boxed{495} \]