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

We are to compute \( BA \) (i.e., multiply matrix \( B \) with \( A \)).
Let’s perform matrix multiplication row-by-column.

Step 1: Multiply Row 1 of \( B \) with all columns of \( A \):
\( R_1 \cdot C_1 = 1 \cdot 2 + 1 \cdot 3 + (-5) \cdot 0 = 2 + 3 + 0 = 5 \)
\( R_1 \cdot C_2 = 1 \cdot 0 + 1 \cdot 4 + (-5) \cdot 2 = 0 + 4 - 10 = -6 \)
\( R_1 \cdot C_3 = 1 \cdot 1 + 1 \cdot 5 + (-5) \cdot 3 = 1 + 5 - 15 = -9 \)

Step-by-step computation:

\[ B = \begin{bmatrix} 1 & 1 & -5 \\ -5 & 1 & -5 \\ 1 & -2 & 4 \end{bmatrix}, \quad A = \begin{bmatrix} 2 & 0 & 1 \\ 3 & 4 & 5 \\ 0 & 2 & 3 \end{bmatrix} \]
Row 1 of \( B \) × Columns of \( A \):
Row: \([1, 1, -5]\)
Column 1: \([2, 3, 0] \Rightarrow 1 \cdot 2 + 1 \cdot 3 + (-5) \cdot 0 = 5\)
Column 2: \([0, 4, 2] \Rightarrow 1 \cdot 0 + 1 \cdot 4 + (-5) \cdot 2 = -6\)
Column 3: \([1, 5, 3] \Rightarrow 1 \cdot 1 + 1 \cdot 5 + (-5) \cdot 3 = -9\)
Row 1 of \( BA \): \([5, -6, -9]\)

Row 2 of \( B \) × Columns of \( A \):
Row: \([-5, 1, -5]\)
C1: \(-5 \cdot 2 + 1 \cdot 3 + (-5) \cdot 0 = -10 + 3 + 0 = -7\)
C2: \(-5 \cdot 0 + 1 \cdot 4 + (-5) \cdot 2 = 0 + 4 - 10 = -6\)
C3: \(-5 \cdot 1 + 1 \cdot 5 + (-5) \cdot 3 = -5 + 5 - 15 = -15\)
Row 2 of \( BA \): \([-7, -6, -15]\)

Row 3 of \( B \) × Columns of \( A \):
Row: \([1, -2, 4]\)
C1: \(1 \cdot 2 + (-2) \cdot 3 + 4 \cdot 0 = 2 - 6 + 0 = -4\)
C2: \(1 \cdot 0 + (-2) \cdot 4 + 4 \cdot 2 = 0 - 8 + 8 = 0\)
C3: \(1 \cdot 1 + (-2) \cdot 5 + 4 \cdot 3 = 1 - 10 + 12 = 3\)
Row 3 of \( BA \): \([-4, 0, 3]\)

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