name:
date:
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unit 10: systems and matrices
homework 5: multiplying matrices
** this is a 2 - page document! **
directions: find each product, if possible.
1. (\begin{bmatrix}4&-1&0\8&4&-3end{bmatrix}cdot\begin{bmatrix}-2\5\1end{bmatrix})
2. (\begin{bmatrix}3&-5\7&1end{bmatrix}cdot\begin{bmatrix}2&2&-4&-7\8&1&3&0end{bmatrix})
3. (\begin{bmatrix}7&11&2\-6&0&-9\10&-4&5end{bmatrix}cdot\begin{bmatrix}5\-3\-1end{bmatrix})
4. (\begin{bmatrix}8\-3\12\-7end{bmatrix}cdot\begin{bmatrix}2&5end{bmatrix})
5. (\begin{bmatrix}-5&-3\-6&7end{bmatrix}cdot\begin{bmatrix}0&3\-7&9\-1&13end{bmatrix})
6. (\begin{bmatrix}4\-7end{bmatrix}cdot\begin{bmatrix}8&-9&6end{bmatrix})
7. (\begin{bmatrix}10&1&6\-4&-1&9end{bmatrix}cdot\begin{bmatrix}12&5\-2&3\-2&-4end{bmatrix})
8. (\begin{bmatrix}7&-8&2&3end{bmatrix}cdot\begin{bmatrix}-5&1\0&-2\3&-3\7&1end{bmatrix})
9. (\begin{bmatrix}2&-5\-1&7end{bmatrix}cdot\begin{bmatrix}11&-6\0&8end{bmatrix}-\begin{bmatrix}3&5\4&-9end{bmatrix})
10. (\begin{bmatrix}14&0\-8&6\3&-2end{bmatrix}+\begin{bmatrix}5\-1\-7end{bmatrix}cdot\begin{bmatrix}3&11end{bmatrix})

Question

Accepted Answer

Let's solve each matrix multiplication problem step by step. We'll check if multiplication is possible (number of columns in first matrix = number of rows in second matrix) and then perform the multiplication.

### Problem 1:
Matrix 1: $\begin{bmatrix} 4 & -1 & 0 \ 8 & 4 & -3 \end{bmatrix}$ (2 rows, 3 columns)
Matrix 2: $\begin{bmatrix} -2 \ 5 \ 1 \end{bmatrix}$ (3 rows, 1 column)
Since 3 (columns of first) = 3 (rows of second), multiplication is possible. Result will be 2x1 matrix.

#### Step 1: Multiply first row of first matrix with column of second matrix
$4(-2) + (-1)(5) + 0(1) = -8 -5 + 0 = -13$

#### Step 2: Multiply second row of first matrix with column of second matrix
$8(-2) + 4(5) + (-3)(1) = -16 + 20 -3 = 1$

# Answer: $\begin{bmatrix} -13 \ 1 \end{bmatrix}$

### Problem 2:
Matrix 1: $\begin{bmatrix} 3 & -5 \ 7 & 1 \end{bmatrix}$ (2 rows, 2 columns)
Matrix 2: $\begin{bmatrix} 2 & 2 & -4 & -7 \ 8 & 1 & 3 & 0 \end{bmatrix}$ (2 rows, 4 columns)
Since 2 (columns of first) = 2 (rows of second), multiplication is possible. Result will be 2x4 matrix.

#### Step 1: First row, first column: $3(2) + (-5)(8) = 6 -40 = -34$
First row, second column: $3(2) + (-5)(1) = 6 -5 = 1$
First row, third column: $3(-4) + (-5)(3) = -12 -15 = -27$
First row, fourth column: $3(-7) + (-5)(0) = -21 + 0 = -21$

#### Step 2: Second row, first column: $7(2) + 1(8) = 14 + 8 = 22$
Second row, second column: $7(2) + 1(1) = 14 + 1 = 15$
Second row, third column: $7(-4) + 1(3) = -28 + 3 = -25$
Second row, fourth column: $7(-7) + 1(0) = -49 + 0 = -49$

# Answer: $\begin{bmatrix} -34 & 1 & -27 & -21 \ 22 & 15 & -25 & -49 \end{bmatrix}$

### Problem 3:
Matrix 1: $\begin{bmatrix} 7 & 11 & 2 \ -6 & 0 & -9 \ 10 & -4 & 5 \end{bmatrix}$ (3 rows, 3 columns)
Matrix 2: $\begin{bmatrix} 5 \ -3 \ -1 \end{bmatrix}$ (3 rows, 1 column)
Since 3 (columns of first) = 3 (rows of second), multiplication is possible. Result will be 3x1 matrix.

#### Step 1: First row: $7(5) + 11(-3) + 2(-1) = 35 -33 -2 = 0$

#### Step 2: Second row: $-6(5) + 0(-3) + (-9)(-1) = -30 + 0 + 9 = -21$

#### Step 3: Third row: $10(5) + (-4)(-3) + 5(-1) = 50 + 12 -5 = 57$

# Answer: $\begin{bmatrix} 0 \ -21 \ 57 \end{bmatrix}$

### Problem 4:
Matrix 1: $\begin{bmatrix} 8 \ -3 \ 12 \ -7 \end{bmatrix}$ (4 rows, 1 column)
Matrix 2: $\begin{bmatrix} 2 & 5 \end{bmatrix}$ (1 row, 2 columns)
Since 1 (columns of first) = 1 (rows of second), multiplication is possible. Result will be 4x2 matrix.

#### Step 1: First row: $8(2) = 16$, $8(5) = 40$

#### Step 2: Second row: $-3(2) = -6$, $-3(5) = -15$

#### Step 3: Third row: $12(2) = 24$, $12(5) = 60$

#### Step 4: Fourth row: $-7(2) = -14$, $-7(5) = -35$

# Answer: $\begin{bmatrix} 16 & 40 \ -6 & -15 \ 24 & 60 \ -14 & -35 \end{bmatrix}$

### Problem 5:
Matrix 1: $\begin{bmatrix} -5 & -3 \ -6 & 7 \end{bmatrix}$ (2 rows, 2 columns)
Matrix 2: $\begin{bmatrix} 0 & 3 \ -7 & 9 \ -1 & 13 \end{bmatrix}$ (3 rows, 2 columns)
Number of columns in first matrix (2) ≠ number of rows in second matrix (3). **Multiplication is not possible.**

# Answer: Not possible (2 columns ≠ 3 rows)

### Problem 6:
Matrix 1: $\begin{bmatrix} 4 \ -7 \end{bmatrix}$ (2 rows, 1 column)
Matrix 2: $\begin{bmatrix} 8 & -9 & 6 \end{bmatrix}$ (1 row, 3 columns)
Since 1 (columns of first) = 1 (rows of second), multiplication is possible. Result will be 2x3 matrix.

#### Step 1: First row: $4(8) = 32$, $4(-9) = -36$, $4(6) = 24$

#### Step 2: Second row: $-7(8) = -56$, $-7(-9) = 63$, $-7(6) = -42$

# Answer: $\begin{bmatrix} 32 & -36 & 24 \ -56 & 63 & -42 \end{bmatrix}$

### Problem 7:
Matrix 1: $\begin{bmatrix} 10 & 1 & 6 \ -4 & -1 & 9 \end{bmatrix}$ (2 rows, 3 columns)
Matrix 2: $\begin{bmatrix} 12 & 5 \ -2 & 3 \ -2 & -4 \end{bmatrix}$ (3 rows, 2 columns)
Since 3 (columns of first) = 3 (rows of second), multiplication is possible. Result will be 2x2 matrix.

#### Step 1: First row, first column: $10(12) + 1(-2) + 6(-2) = 120 -2 -12 = 106$
First row, second column: $10(5) + 1(3) + 6(-4) = 50 + 3 -24 = 29$

#### Step 2: Second row, first column: $-4(12) + (-1)(-2) + 9(-2) = -48 + 2 -18 = -64$
Second row, second column: $-4(5) + (-1)(3) + 9(-4) = -20 -3 -36 = -59$

# Answer: $\begin{bmatrix} 106 & 29 \ -64 & -59 \end{bmatrix}$

### Problem 8:
Matrix 1: $\begin{bmatrix} 7 & -8 & 2 & 3 \end{bmatrix}$ (1 row, 4 columns)
Matrix 2: $\begin{bmatrix} -5 & 1 \ 0 & -2 \ 3 & -3 \ 7 & 1 \end{bmatrix}$ (4 rows, 2 columns)
Since 4 (columns of first) = 4 (rows of second), multiplication is possible. Result will be 1x2 matrix.

#### Step 1: First element: $7(-5) + (-8)(0) + 2(3) + 3(7) = -35 + 0 + 6 + 21 = -8$
Second element: $7(1) + (-8)(-2) + 2(-3) + 3(1) = 7 + 16 -6 + 3 = 20$

# Answer: $\begin{bmatrix} -8 & 20 \end{bmatrix}$

### Problem 9:
First, multiply $\begin{bmatrix} 2 & -5 \ -1 & 7 \end{bmatrix}$ and $\begin{bmatrix} 11 & -6 \ 0 & 8 \end{bmatrix}$, then multiply the result by $\begin{bmatrix} 3 & 5 \ 4 & -9 \end{bmatrix}$.

#### Step 1: Multiply first two matrices (2x2 * 2x2 = 2x2)
- First row, first column: $2(11) + (-5)(0) = 22$
- First row, second column: $2(-6) + (-5)(8) = -12 -40 = -52$
- Second row, first column: $-1(11) + 7(0) = -11$
- Second row, second column: $-1(-6) + 7(8) = 6 + 56 = 62$
Result: $\begin{bmatrix} 22 & -52 \ -11 & 62 \end{bmatrix}$

#### Step 2: Multiply this result by $\begin{bmatrix} 3 & 5 \ 4 & -9 \end{bmatrix}$ (2x2 * 2x2 = 2x2)
- First row, first column: $22(3) + (-52)(4) = 66 -208 = -142$
- First row, second column: $22(5) + (-52)(-9) = 110 + 468 = 578$
- Second row, first column: $-11(3) + 62(4) = -33 + 248 = 215$
- Second row, second column: $-11(5) + 62(-9) = -55 -558 = -613$

# Answer: $\begin{bmatrix} -142 & 578 \ 215 & -613 \end{bmatrix}$

### Problem 10:
First, multiply $\begin{bmatrix} 14 & 0 \ -8 & 6 \ 3 & -2 \end{bmatrix}$ and $\begin{bmatrix} 5 \ -1 \ -7 \end{bmatrix}$? Wait, no—wait, the problem is $\begin{bmatrix} 14 & 0 \ -8 & 6 \ 3 & -2 \end{bmatrix} + \begin{bmatrix} 5 \ -1 \ -7 \end{bmatrix} \cdot \begin{bmatrix} 3 & 11 \end{bmatrix}$? Wait, no, let's parse the problem:

Wait, the problem is: $\begin{bmatrix} 14 & 0 \ -8 & 6 \ 3 & -2 \end{bmatrix} + \begin{bmatrix} 5 \ -1 \ -7 \end{bmatrix} \cdot \begin{bmatrix} 3 & 11 \end{bmatrix}$

First, multiply $\begin{bmatrix} 5 \ -1 \ -7 \end{bmatrix}$ (3x1) and $\begin{bmatrix} 3 & 11 \end{bmatrix}$ (1x2) → 3x2 matrix. Then add to $\begin{bmatrix} 14 & 0 \ -8 & 6 \ 3 & -2 \end{bmatrix}$ (3x2).

#### Step 1: Multiply $\begin{bmatrix} 5 \ -1 \ -7 \end{bmatrix}$ and $\begin{bmatrix} 3 & 11 \end{bmatrix}$
- First row: $5(3) = 15$, $5(11) = 55$
- Second row: $-1(3) = -3$, $-1(11) = -11$
- Third row: $-7(3) = -21$, $-7(11) = -77$
Result: $\begin{bmatrix} 15 & 55 \ -3 & -11 \ -21 & -77 \end{bmatrix}$

#### Step 2: Add to $\begin{bmatrix} 14 & 0 \ -8 & 6 \ 3 & -2 \end{bmatrix}$
- First row: $14 + 15 = 29$, $0 + 55 = 55$
- Second row: $-8 + (-3) = -11$, $6 + (-11) = -5$
- Third row: $3 + (-21) = -18$, $-2 + (-77) = -79$

# Answer: $\begin{bmatrix} 29 & 55 \ -11 & -5 \ -18 & -79 \end{bmatrix}$

### Final Answers:
1. $\boldsymbol{\begin{bmatrix} -13 \ 1 \end{bmatrix}}$
2. $\boldsymbol{\begin{bmatrix} -34 & 1 & -27 & -21 \ 22 & 15 & -25 & -49 \end{bmatrix}}$
3. $\boldsymbol{\begin{bmatrix} 0 \ -21 \ 57 \end{bmatrix}}$
4. $\boldsymbol{\begin{bmatrix} 16 & 40 \ -6 & -15 \ 24 & 60 \ -14 & -35 \end{bmatrix}}$
5. Not possible
6. $\boldsymbol{\begin{bmatrix} 32 & -36 & 24 \ -56 & 63 & -42 \end{bmatrix}}$
7. $\boldsymbol{\begin{bmatrix} 106 & 29 \ -64 & -59 \end{bmatrix}}$
8. $\boldsymbol{\begin{bmatrix} -8 & 20 \end{bmatrix}}$
9. $\boldsymbol{\begin{bmatrix} -142 & 578 \ 215 & -613 \end{bmatrix}}$
10. $\boldsymbol{\begin{bmatrix} 29 & 55 \ -11 & -5 \ -18 & -79 \end{bmatrix}}$

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step 1: Multiply first row of first matrix with column of second matrix

Step 2: Multiply second row of first matrix with column of second matrix

Answer:

Problem 2:

Step 1: First row, first column: $3(2) + (-5)(8) = 6 -40 = -34$

Step 2: Second row, first column: $7(2) + 1(8) = 14 + 8 = 22$