Question

Find the product of the following two matrices.
open square bracket,
begin matrix
row 1, column 1, 4
row 1, column 2, 0
row 2, column 1, 0
row 2, column 2, 3
end matrix
, close square bracket, open square bracket,
begin matrix
row 1, column 1, minus, 1
row 1, column 2, minus, 2
row 2, column 1, minus, 1
row 2, column 2, 1
end matrix
, close square bracket
[
4
0
​

0
3
​
][
−1
−1
​

−2
1
​
]

Answer 1

To find the product of the two given matrices, we need to perform matrix multiplication. Let's denote the first matrix as \( A \) and the second matrix as \( B \):

\[
A = \begin{pmatrix}
4 & 0 \\
0 & 3
\end{pmatrix}, \quad B = \begin{pmatrix}
-1 & -2 \\
-1 & 1
\end{pmatrix}
\]

The product of two matrices \( A \) and \( B \) (resulting in matrix \( C = AB \)) is calculated by taking the dot product of rows of \( A \) with columns of \( B \).

The resulting matrix \( C \) will have the following entries:

- \( C_{11} \): First row of \( A \) dot with first column of \( B \)
- \( C_{12} \): First row of \( A \) dot with second column of \( B \)
- \( C_{21} \): Second row of \( A \) dot with first column of \( B \)
- \( C_{22} \): Second row of \( A \) dot with second column of \( B \)

Now we will compute each entry:

1. \( C_{11} = 4 \cdot (-1) + 0 \cdot (-1) = -4 + 0 = -4 \)
2. \( C_{12} = 4 \cdot (-2) + 0 \cdot 1 = -8 + 0 = -8 \)
3. \( C_{21} = 0 \cdot (-1) + 3 \cdot (-1) = 0 - 3 = -3 \)
4. \( C_{22} = 0 \cdot (-2) + 3 \cdot 1 = 0 + 3 = 3 \)

Putting it all together, we find that:

\[
C = AB = \begin{pmatrix}
-4 & -8 \\
-3 & 3
\end{pmatrix}
\]

Thus, the product of the two matrices is:

\[
\begin{pmatrix}
-4 & -8 \\
-3 & 3
\end{pmatrix}
\]