Matrix multiplication practice worksheet with six problems.

Problem: Matrix Multiplication

The task is to simplify the given matrix multiplication problems. If the expression is not defined, write "undefined."

#### Matrix Multiplication Rules Recap:
1. To multiply two matrices \( A \) and \( B \), the number of columns in \( A \) must equal the number of rows in \( B \).
2. The resulting matrix will have the same number of rows as \( A \) and the same number of columns as \( B \).
3. Each element in the resulting matrix is computed as the dot product of a row from \( A \) and a column from \( B \).

---

Problem 1:

\[
\begin{bmatrix}
-6 & 1 & -1 \\
-5 & 0 & -2
\end{bmatrix}
\cdot
\begin{bmatrix}
-3 \\
-5
\end{bmatrix}
\]

#### Step 1: Check Dimensions
- Matrix \( A \) is \( 2 \times 3 \).
- Matrix \( B \) is \( 2 \times 1 \).
- The number of columns in \( A \) (3) does not match the number of rows in \( B \) (2). Therefore, this multiplication is undefined.

#### Answer for Problem 1:
\[
\boxed{\text{undefined}}
\]

---

Problem 2:

\[
\begin{bmatrix}
5 & 0 & -3 \\
0 & -3 & -1 \\
-1 & 1 & 4 \\
4 & -3 & -6
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & -4 \\
3 & -2 \\
-1 & 6
\end{bmatrix}
\]

#### Step 1: Check Dimensions
- Matrix \( A \) is \( 4 \times 3 \).
- Matrix \( B \) is \( 3 \times 2 \).
- The number of columns in \( A \) (3) matches the number of rows in \( B \) (3). The result will be a \( 4 \times 2 \) matrix.

#### Step 2: Compute the Result
The resulting matrix \( C \) is computed as:
\[
C_{ij} = \sum_{k=1}^3 A_{ik} B_{kj}
\]

##### Row 1, Column 1:
\[
C_{11} = (5)(1) + (0)(3) + (-3)(-1) = 5 + 0 + 3 = 8
\]

##### Row 1, Column 2:
\[
C_{12} = (5)(-4) + (0)(-2) + (-3)(6) = -20 + 0 - 18 = -38
\]

##### Row 2, Column 1:
\[
C_{21} = (0)(1) + (-3)(3) + (-1)(-1) = 0 - 9 + 1 = -8
\]

##### Row 2, Column 2:
\[
C_{22} = (0)(-4) + (-3)(-2) + (-1)(6) = 0 + 6 - 6 = 0
\]

##### Row 3, Column 1:
\[
C_{31} = (-1)(1) + (1)(3) + (4)(-1) = -1 + 3 - 4 = -2
\]

##### Row 3, Column 2:
\[
C_{32} = (-1)(-4) + (1)(-2) + (4)(6) = 4 - 2 + 24 = 26
\]

##### Row 4, Column 1:
\[
C_{41} = (4)(1) + (-3)(3) + (-6)(-1) = 4 - 9 + 6 = 1
\]

##### Row 4, Column 2:
\[
C_{42} = (4)(-4) + (-3)(-2) + (-6)(6) = -16 + 6 - 36 = -46
\]

#### Final Result for Problem 2:
\[
\begin{bmatrix}
8 & -38 \\
-8 & 0 \\
-2 & 26 \\
1 & -46
\end{bmatrix}
\]

---

Problem 3:

\[
\begin{bmatrix}
-5 & 2 \\
-4 & 4 \\
3 & 4 \\
0 & -2
\end{bmatrix}
\cdot
\begin{bmatrix}
5 & 1 & 1 \\
-2 & 0 & 2
\end{bmatrix}
\]

#### Step 1: Check Dimensions
- Matrix \( A \) is \( 4 \times 2 \).
- Matrix \( B \) is \( 2 \times 3 \).
- The number of columns in \( A \) (2) matches the number of rows in \( B \) (2). The result will be a \( 4 \times 3 \) matrix.

#### Step 2: Compute the Result
The resulting matrix \( C \) is computed as:
\[
C_{ij} = \sum_{k=1}^2 A_{ik} B_{kj}
\]

##### Row 1, Column 1:
\[
C_{11} = (-5)(5) + (2)(-2) = -25 - 4 = -29
\]

##### Row 1, Column 2:
\[
C_{12} = (-5)(1) + (2)(0) = -5 + 0 = -5
\]

##### Row 1, Column 3:
\[
C_{13} = (-5)(1) + (2)(2) = -5 + 4 = -1
\]

##### Row 2, Column 1:
\[
C_{21} = (-4)(5) + (4)(-2) = -20 - 8 = -28
\]

##### Row 2, Column 2:
\[
C_{22} = (-4)(1) + (4)(0) = -4 + 0 = -4
\]

##### Row 2, Column 3:
\[
C_{23} = (-4)(1) + (4)(2) = -4 + 8 = 4
\]

##### Row 3, Column 1:
\[
C_{31} = (3)(5) + (4)(-2) = 15 - 8 = 7
\]

##### Row 3, Column 2:
\[
C_{32} = (3)(1) + (4)(0) = 3 + 0 = 3
\]

##### Row 3, Column 3:
\[
C_{33} = (3)(1) + (4)(2) = 3 + 8 = 11
\]

##### Row 4, Column 1:
\[
C_{41} = (0)(5) + (-2)(-2) = 0 + 4 = 4
\]

##### Row 4, Column 2:
\[
C_{42} = (0)(1) + (-2)(0) = 0 + 0 = 0
\]

##### Row 4, Column 3:
\[
C_{43} = (0)(1) + (-2)(2) = 0 - 4 = -4
\]

#### Final Result for Problem 3:
\[
\begin{bmatrix}
-29 & -5 & -1 \\
-28 & -4 & 4 \\
7 & 3 & 11 \\
4 & 0 & -4
\end{bmatrix}
\]

---

Problem 4:

\[
\begin{bmatrix}
0 & -3 \\
-5 & 2 \\
-1 & 3
\end{bmatrix}
\cdot
\begin{bmatrix}
2 & 3 \\
0 & -2 \\
\end{bmatrix}
\]

#### Step 1: Check Dimensions
- Matrix \( A \) is \( 3 \times 2 \).
- Matrix \( B \) is \( 2 \times 2 \).
- The number of columns in \( A \) (2) matches the number of rows in \( B \) (2). The result will be a \( 3 \times 2 \) matrix.

#### Step 2: Compute the Result
The resulting matrix \( C \) is computed as:
\[
C_{ij} = \sum_{k=1}^2 A_{ik} B_{kj}
\]

##### Row 1, Column 1:
\[
C_{11} = (0)(2) + (-3)(0) = 0 + 0 = 0
\]

##### Row 1, Column 2:
\[
C_{12} = (0)(3) + (-3)(-2) = 0 + 6 = 6
\]

##### Row 2, Column 1:
\[
C_{21} = (-5)(2) + (2)(0) = -10 + 0 = -10
\]

##### Row 2, Column 2:
\[
C_{22} = (-5)(3) + (2)(-2) = -15 - 4 = -19
\]

##### Row 3, Column 1:
\[
C_{31} = (-1)(2) + (3)(0) = -2 + 0 = -2
\]

##### Row 3, Column 2:
\[
C_{32} = (-1)(3) + (3)(-2) = -3 - 6 = -9
\]

#### Final Result for Problem 4:
\[
\begin{bmatrix}
0 & 6 \\
-10 & -19 \\
-2 & -9
\end{bmatrix}
\]

---

Problem 5:

\[
\begin{bmatrix}
2 \\
-2 \\
5
\end{bmatrix}
\cdot
\begin{bmatrix}
4 & -3
\end{bmatrix}
\]

#### Step 1: Check Dimensions
- Matrix \( A \) is \( 3 \times 1 \).
- Matrix \( B \) is \( 1 \times 2 \).
- The number of columns in \( A \) (1) matches the number of rows in \( B \) (1). The result will be a \( 3 \times 2 \) matrix.

#### Step 2: Compute the Result
The resulting matrix \( C \) is computed as:
\[
C_{ij} = A_{i1} B_{1j}
\]

##### Row 1, Column 1:
\[
C_{11} = (2)(4) = 8
\]

##### Row 1, Column 2:
\[
C_{12} = (2)(-3) = -6
\]

##### Row 2, Column 1:
\[
C_{21} = (-2)(4) = -8
\]

##### Row 2, Column 2:
\[
C_{22} = (-2)(-3) = 6
\]

##### Row 3, Column 1:
\[
C_{31} = (5)(4) = 20
\]

##### Row 3, Column 2:
\[
C_{32} = (5)(-3) = -15
\]

#### Final Result for Problem 5:
\[
\begin{bmatrix}
8 & -6 \\
-8 & 6 \\
20 & -15
\end{bmatrix}
\]

---

Problem 6:

\[
\begin{bmatrix}
5 & -4 \\
-2 & -2 \\
-1 & 6
\end{bmatrix}
\cdot
\begin{bmatrix}
-4 & -1 \\
6 & -4
\end{bmatrix}
\]

#### Step 1: Check Dimensions
- Matrix \( A \) is \( 3 \times 2 \).
- Matrix \( B \) is \( 2 \times 2 \).
- The number of columns in \( A \) (2) matches the number of rows in \( B \) (2). The result will be a \( 3 \times 2 \) matrix.

#### Step 2: Compute the Result
The resulting matrix \( C \) is computed as:
\[
C_{ij} = \sum_{k=1}^2 A_{ik} B_{kj}
\]

##### Row 1, Column 1:
\[
C_{11} = (5)(-4) + (-4)(6) = -20 - 24 = -44
\]

##### Row 1, Column 2:
\[
C_{12} = (5)(-1) + (-4)(-4) = -5 + 16 = 11
\]

##### Row 2, Column 1:
\[
C_{21} = (-2)(-4) + (-2)(6) = 8 - 12 = -4
\]

##### Row 2, Column 2:
\[
C_{22} = (-2)(-1) + (-2)(-4) = 2 + 8 = 10
\]

##### Row 3, Column 1:
\[
C_{31} = (-1)(-4) + (6)(6) = 4 + 36 = 40
\]

##### Row 3, Column 2:
\[
C_{32} = (-1)(-1) + (6)(-4) = 1 - 24 = -23
\]

#### Final Result for Problem 6:
\[
\begin{bmatrix}
-44 & 11 \\
-4 & 10 \\
40 & -23
\end{bmatrix}
\]

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Final Answers:

1. \(\boxed{\text{undefined}}\)
2. \(\boxed{\begin{bmatrix} 8 & -38 \\ -8 & 0 \\ -2 & 26 \\ 1 & -46 \end{bmatrix}}\)
3. \(\boxed{\begin{bmatrix} -29 & -5 & -1 \\ -28 & -4 & 4 \\ 7 & 3 & 11 \\ 4 & 0 & -4 \end{bmatrix}}\)
4. \(\boxed{\begin{bmatrix} 0 & 6 \\ -10 & -19 \\ -2 & -9 \end{bmatrix}}\)
5. \(\boxed{\begin{bmatrix} 8 & -6 \\ -8 & 6 \\ 20 & -15 \end{bmatrix}}\)
6. \(\boxed{\begin{bmatrix} -44 & 11 \\ -4 & 10 \\ 40 & -23 \end{bmatrix}}\)