Question

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

0
−4
​
][
0
−3
​

1
−1
​
]

Answer 1

To find the product of the two matrices, we will multiply them according to the rules of matrix multiplication.

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

The product \( AB \) will be calculated to form a new matrix \( C \):

\[
C = AB = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix}
\]

Where:
- \( c_{11} \) is the dot product of the first row of \( A \) and the first column of \( B \).
- \( c_{12} \) is the dot product of the first row of \( A \) and the second column of \( B \).
- \( c_{21} \) is the dot product of the second row of \( A \) and the first column of \( B \).
- \( c_{22} \) is the dot product of the second row of \( A \) and the second column of \( B \).

Let's calculate each element in \( C \):

1. **Calculating \( c_{11} \)**:

\[
c_{11} = (-2)(0) + (0)(-3) = 0 + 0 = 0
\]

2. **Calculating \( c_{12} \)**:

\[
c_{12} = (-2)(1) + (0)(-1) = -2 + 0 = -2
\]

3. **Calculating \( c_{21} \)**:

\[
c_{21} = (3)(0) + (-4)(-3) = 0 + 12 = 12
\]

4. **Calculating \( c_{22} \)**:

\[
c_{22} = (3)(1) + (-4)(-1) = 3 + 4 = 7
\]

Putting it all together, we get the product matrix \( C \):

\[
C = \begin{pmatrix} 0 & -2 \\ 12 & 7 \end{pmatrix}
\]

Thus, the product of the two matrices is:

\[
\begin{pmatrix} 0 & -2 \\ 12 & 7 \end{pmatrix}
\]