Let's solve each matrix multiplication problem step by step. Matrix multiplication involves multiplying rows of the first matrix by columns of the second matrix.
---
1.
$$
\begin{bmatrix}
3 & -5 \\
-4 & 2
\end{bmatrix}
\times
\begin{bmatrix}
-2 & 0 \\
5 & -6
\end{bmatrix}
$$
We compute each element of the resulting $2 \times 2$ matrix:
-
Row 1, Column 1:
$3 \cdot (-2) + (-5) \cdot 5 = -6 - 25 = -31$
-
Row 1, Column 2:
$3 \cdot 0 + (-5) \cdot (-6) = 0 + 30 = 30$
-
Row 2, Column 1:
$-4 \cdot (-2) + 2 \cdot 5 = 8 + 10 = 18$
-
Row 2, Column 2:
$-4 \cdot 0 + 2 \cdot (-6) = 0 - 12 = -12$
✔ Result:
$$
\begin{bmatrix}
-31 & 30 \\
18 & -12
\end{bmatrix}
$$
---
2.
$$
\begin{bmatrix}
-2 & 7 \\
4 & -5
\end{bmatrix}
\times
\begin{bmatrix}
-7 & 9 \\
4 & -3
\end{bmatrix}
$$
-
Row 1, Col 1:
$(-2)(-7) + (7)(4) = 14 + 28 = 42$
-
Row 1, Col 2:
$(-2)(9) + (7)(-3) = -18 - 21 = -39$
-
Row 2, Col 1:
$(4)(-7) + (-5)(4) = -28 - 20 = -48$
-
Row 2, Col 2:
$(4)(9) + (-5)(-3) = 36 + 15 = 51$
✔ Result:
$$
\begin{bmatrix}
42 & -39 \\
-48 & 51
\end{bmatrix}
$$
---
3.
$$
\begin{bmatrix}
-9 & 2 \\
8 & -1 \\
4 & -3
\end{bmatrix}
\times
\begin{bmatrix}
-2 & 5 \\
3 & -6
\end{bmatrix}
$$
This is a $3 \times 2$ matrix times a $2 \times 2$ matrix → result is $3 \times 2$.
-
Row 1, Col 1:
$(-9)(-2) + (2)(3) = 18 + 6 = 24$
-
Row 1, Col 2:
$(-9)(5) + (2)(-6) = -45 - 12 = -57$
-
Row 2, Col 1:
$(8)(-2) + (-1)(3) = -16 - 3 = -19$
-
Row 2, Col 2:
$(8)(5) + (-1)(-6) = 40 + 6 = 46$
-
Row 3, Col 1:
$(4)(-2) + (-3)(3) = -8 - 9 = -17$
-
Row 3, Col 2:
$(4)(5) + (-3)(-6) = 20 + 18 = 38$
✔ Result:
$$
\begin{bmatrix}
24 & -57 \\
-19 & 46 \\
-17 & 38
\end{bmatrix}
$$
---
4.
$$
\begin{bmatrix}
2 & -1 \\
5 & 8 \\
4 & 0
\end{bmatrix}
\times
\begin{bmatrix}
3 & 2 \\
-4 & -7
\end{bmatrix}
$$
Again: $3 \times 2$ × $2 \times 2$ → $3 \times 2$
-
Row 1, Col 1:
$(2)(3) + (-1)(-4) = 6 + 4 = 10$
-
Row 1, Col 2:
$(2)(2) + (-1)(-7) = 4 + 7 = 11$
-
Row 2, Col 1:
$(5)(3) + (8)(-4) = 15 - 32 = -17$
-
Row 2, Col 2:
$(5)(2) + (8)(-7) = 10 - 56 = -46$
-
Row 3, Col 1:
$(4)(3) + (0)(-4) = 12 + 0 = 12$
-
Row 3, Col 2:
$(4)(2) + (0)(-7) = 8 + 0 = 8$
✔ Result:
$$
\begin{bmatrix}
10 & 11 \\
-17 & -46 \\
12 & 8
\end{bmatrix}
$$
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✔ Final Answers:
1. $\begin{bmatrix} -31 & 30 \\ 18 & -12 \end{bmatrix}$
2. $\begin{bmatrix} 42 & -39 \\ -48 & 51 \end{bmatrix}$
3. $\begin{bmatrix} 24 & -57 \\ -19 & 46 \\ -17 & 38 \end{bmatrix}$
4. $\begin{bmatrix} 10 & 11 \\ -17 & -46 \\ 12 & 8 \end{bmatrix}$
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Parent Tip: Review the logic above to help your child master the concept of free algebra 2 worksheet.
