Question

Given \( A = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix} \), find \( BA \).

Hint:

When multiplying matrices, perform row-by-column dot product. Make sure dimensions match: For $m \times n$ × $n \times p$ matrices, result is $m \times p$.

Answer 1

The objective is to compute the product \( BA \). Matrix multiplication will be performed row-by-column.

The matrices involved are:
\[ B = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix}, \quadA = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} \]
Computation for Row 1 of \( BA \):
Row 1 of \( B \) is \([1, 1, -5]\).
Multiplying with Column 1 of \( A \): \([2, 3, 0]\) results in \(1 \cdot 2 + 1 \cdot 3 + (-5) \cdot 0 = 2 + 3 + 0 = 5\).
Multiplying with Column 2 of \( A \): \([0, 4, 2]\) results in \(1 \cdot 0 + 1 \cdot 4 + (-5) \cdot 2 = 0 + 4 - 10 = -6\).
Multiplying with Column 3 of \( A \): \([1, 5, 3]\) results in \(1 \cdot 1 + 1 \cdot 5 + (-5) \cdot 3 = 1 + 5 - 15 = -9\).
Thus, Row 1 of \( BA \) is \([5, -6, -9]\).

Computation for Row 2 of \( BA \):
Row 2 of \( B \) is \([-5, 1, -5]\).
Multiplying with Column 1 of \( A \): \([2, 3, 0]\) results in \(-5 \cdot 2 + 1 \cdot 3 + (-5) \cdot 0 = -10 + 3 + 0 = -7\).
Multiplying with Column 2 of \( A \): \([0, 4, 2]\) results in \(-5 \cdot 0 + 1 \cdot 4 + (-5) \cdot 2 = 0 + 4 - 10 = -6\).
Multiplying with Column 3 of \( A \): \([1, 5, 3]\) results in \(-5 \cdot 1 + 1 \cdot 5 + (-5) \cdot 3 = -5 + 5 - 15 = -15\).
Thus, Row 2 of \( BA \) is \([-7, -6, -15]\).

Computation for Row 3 of \( BA \):
Row 3 of \( B \) is \([1, -2, 4]\).
Multiplying with Column 1 of \( A \): \([2, 3, 0]\) results in \(1 \cdot 2 + (-2) \cdot 3 + 4 \cdot 0 = 2 - 6 + 0 = -4\).
Multiplying with Column 2 of \( A \): \([0, 4, 2]\) results in \(1 \cdot 0 + (-2) \cdot 4 + 4 \cdot 2 = 0 - 8 + 8 = 0\).
Multiplying with Column 3 of \( A \): \([1, 5, 3]\) results in \(1 \cdot 1 + (-2) \cdot 5 + 4 \cdot 3 = 1 - 10 + 12 = 3\).
Thus, Row 3 of \( BA \) is \([-4, 0, 3]\).

Final Result:
\[ BA = \begin{bmatrix} 5 & -6 & -9 \\ -7 & -6 & -15 \\ -4 & 0 & 3 \end{bmatrix} \]