To perform the matrix multiplication \( AB \) where \( A \) and \( B \) are given matrices, we need to follow the steps for matrix multiplication. The result matrix \( C \) will be of the same dimensions as the row count of \( A \) by the column count of \( B \). Given:
[tex]\[
A = \begin{pmatrix}
7 & 3 \\
-4 & 2 \\
-8 & 8
\end{pmatrix}, \quad B = \begin{pmatrix}
6 & -6 \\
-10 & -7
\end{pmatrix}
\][/tex]
We will multiply row vectors of \( A \) by column vectors of \( B \).
1. Compute the first row of \( C \) (i.e., first row of \( A \) times each column of \( B \)):
- First row of \( A \): \([7, 3]\)
- First column of \( B \): \([6, -10]\)
\( 7 \cdot 6 + 3 \cdot (-10) = 42 - 30 = 12 \)
- Second column of \( B \): \([-6, -7]\)
\( 7 \cdot (-6) + 3 \cdot (-7) = -42 - 21 = -63 \)
So, the first row of \( C \) is \([12, -63]\).
2. Compute the second row of \( C \) (i.e., second row of \( A \) times each column of \( B \)):
- Second row of \( A \): \([-4, 2]\)
- First column of \( B \): \([6, -10]\)
\( -4 \cdot 6 + 2 \cdot (-10) = -24 - 20 = -44 \)
- Second column of \( B \): \([-6, -7]\)
\( -4 \cdot (-6) + 2 \cdot (-7) = 24 - 14 = 10 \)
So, the second row of \( C \) is \([-44, 10]\).
3. Compute the third row of \( C \) (i.e., third row of \( A \) times each column of \( B \)):
- Third row of \( A \): \([-8, 8]\)
- First column of \( B \): \([6, -10]\)
\( -8 \cdot 6 + 8 \cdot (-10) = -48 - 80 = -128 \)
- Second column of \( B \): \([-6, -7]\)
\( -8 \cdot (-6) + 8 \cdot (-7) = 48 - 56 = -8 \)
So, the third row of \( C \) is \([-128, -8]\).
Putting it all together, the resultant matrix \( C \) is:
[tex]\[
C = \begin{pmatrix}
12 & -63 \\
-44 & 10 \\
-128 & -8
\end{pmatrix}
\][/tex]
Therefore, the product \( AB \) is:
[tex]\[
AB = \begin{pmatrix}
12 & -63 \\
-44 & 10 \\
-128 & -8
\end{pmatrix}
\][/tex]
