Question

date 11/2-6
period 1
addition/subtraction/scalar multiplying matrices worksheet
perform the indicated operation, if possible. if not possible, state the reason

1. $\begin{bmatrix} 3&13\\ 6&4 end{bmatrix}-\begin{bmatrix} -2&11\\ -4&0 end{bmatrix}$
2. $\begin{bmatrix} 5&2\\ 8&4\\ -1&8\\ -3&0 end{bmatrix}-\begin{bmatrix} 2&4\\ 3&-2\\ 6&7\\ 7&-5 end{bmatrix}$
3. $\begin{bmatrix} 9&1&4\\ 1&-3&2\\ 8&7&1 end{bmatrix}-\begin{bmatrix} 4&5&-8\\ 5&1&0\\ 6&4&7 end{bmatrix}$
4. $-4 2 3+\begin{bmatrix} -2\\ 0\\ -1 end{bmatrix}$
5. $\begin{bmatrix} 10&-5&7\\ 2&-12&0\\ 8&-4&6 end{bmatrix}+\begin{bmatrix} -7&14&6\\ 0&12&-4\\ 2&7&3 end{bmatrix}$
6. $\begin{bmatrix} 10&-7&14\\ -5&-10&0\\ 9&-3&-7 end{bmatrix}-\begin{bmatrix} -1&-3&8\\ -12&0&6\\ 10&-5&5 end{bmatrix}$

perform the indicated operation

1. $-3\begin{bmatrix} 4&2\\ 3&2 end{bmatrix}$
2. $-2\begin{bmatrix} 3&0&-1\\ 0.5&-6&4\\ 7&-1.25&9 end{bmatrix}-4\begin{bmatrix} 4&1\\ -5&0\\ 1&-3 end{bmatrix}$

Explanation:

Step1: Identify original matrices

Original matrices:

$$\begin{bmatrix} 2 & 13 \\ 6 & 4 \end{bmatrix} - \begin{bmatrix} -2 & 14 \\ -4 & 0 \end{bmatrix}$$

Step2: Subtract corresponding elements

$$\begin{bmatrix} 2-(-2) & 13-14 \\ 6-(-4) & 4-0 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 10 & 4 \end{bmatrix}$$

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Step1: Identify original matrices

Original matrices:

$$\begin{bmatrix} 5 & 21 \\ -1 & 8 \\ -3 & 0 \end{bmatrix} + \begin{bmatrix} -2 & 4 \\ 6 & -6 \\ 7 & -5 \end{bmatrix}$$

Step2: Add corresponding elements

$$\begin{bmatrix} 5+(-2) & 21+4 \\ -1+6 & 8+(-6) \\ -3+7 & 0+(-5) \end{bmatrix} = \begin{bmatrix} 3 & 25 \\ 5 & 2 \\ 4 & -5 \end{bmatrix}$$

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Step1: Identify original matrices

Original matrices:

$$\begin{bmatrix} 9 & 4 & 3 \\ 1 & -2 & 2 \\ 8 & 7 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 5 & -8 \\ 5 & 8 & 0 \\ 6 & 4 & 7 \end{bmatrix}$$

Step2: Subtract corresponding elements

$$\begin{bmatrix} 9-4 & 4-5 & 3-(-8) \\ 1-5 & -2-8 & 2-0 \\ 8-6 & 7-4 & 1-7 \end{bmatrix} = \begin{bmatrix} 5 & -1 & 11 \\ -4 & -10 & 2 \\ 2 & 3 & -6 \end{bmatrix}$$

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Step1: Verify matrix multiplication eligibility

1x3 row matrix

$$\begin{bmatrix} -4 & 2 & 3 \end{bmatrix}$$

and 3x1 column matrix

$$\begin{bmatrix} -2 \\ 0 \\ -1 \end{bmatrix}$$

are compatible.

Step2: Calculate dot product

$(-4)(-2) + (2)(0) + (3)(-1) = 8 + 0 - 3 = 5$

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Step1: Identify original matrices

Original matrices:

$$\begin{bmatrix} 10 & -5 & 7 \\ 2 & -12 & 0 \\ 8 & -4 & 6 \end{bmatrix} + \begin{bmatrix} -7 & 14 & 6 \\ 0 & 12 & -4 \\ 2 & 7 & 3 \end{bmatrix}$$

Step2: Add corresponding elements

$$\begin{bmatrix} 10+(-7) & -5+14 & 7+6 \\ 2+0 & -12+12 & 0+(-4) \\ 8+2 & -4+7 & 6+3 \end{bmatrix} = \begin{bmatrix} 3 & 9 & 13 \\ 2 & 0 & -4 \\ 10 & 3 & 9 \end{bmatrix}$$

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Step1: Identify original matrices

Original matrices:

$$\begin{bmatrix} 10 & -7 & 14 \\ -5 & -10 & 0 \\ 9 & -3 & -7 \end{bmatrix} - \begin{bmatrix} -1 & -3 & 8 \\ -12 & 0 & 6 \\ 10 & -5 & 5 \end{bmatrix}$$

Step2: Subtract corresponding elements

$$\begin{bmatrix} 10-(-1) & -7-(-3) & 14-8 \\ -5-(-12) & -10-0 & 0-6 \\ 9-10 & -3-(-5) & -7-5 \end{bmatrix} = \begin{bmatrix} 11 & -4 & 6 \\ 7 & -10 & -6 \\ -1 & 2 & -12 \end{bmatrix}$$

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Step1: Multiply scalar by each element

Scalar $-3$ times

$$\begin{bmatrix} 4 & 2 \\ 3 & 2 \end{bmatrix}$$

:

$$\begin{bmatrix} -3(4) & -3(2) \\ -3(3) & -3(2) \end{bmatrix} = \begin{bmatrix} -12 & -6 \\ -9 & -6 \end{bmatrix}$$

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Step1: Calculate first scalar multiplication

Scalar $-2$ times

$$\begin{bmatrix} 3 & 0 & -1 \\ 0.5 & -6 & 4 \\ 7 & -1.25 & 9 \end{bmatrix}$$

:

$$\begin{bmatrix} -2(3) & -2(0) & -2(-1) \\ -2(0.5) & -2(-6) & -2(4) \\ -2(7) & -2(-1.25) & -2(9) \end{bmatrix} = \begin{bmatrix} -6 & 0 & 2 \\ -1 & 12 & -8 \\ -14 & 2.5 & -18 \end{bmatrix}$$

Step2: Calculate second scalar multiplication

Scalar $-4$ times

$$\begin{bmatrix} 4 & 1 \\ -5 & 0 \\ 1 & -3 \end{bmatrix}$$

:

$$\begin{bmatrix} -4(4) & -4(1) \\ -4(-5) & -4(0) \\ -4(1) & -4(-3) \end{bmatrix} = \begin{bmatrix} -16 & -4 \\ 20 & 0 \\ -4 & 12 \end{bmatrix}$$

Step3: Subtract the two matrices

$$\begin{bmatrix} -6-(-16) & 0-(-4) & 2-0 \\ -1-20 & 12-0 & -8-0 \\ -14-(-4) & 2.5-0 & -18-12 \end{bmatrix} = \begin{bmatrix} 10 & 4 & 2 \\ -21 & 12 & -8 \\ -10 & 2.5 & -30 \end{bmatrix}$$

Answer:

1.

$$\begin{bmatrix} 4 & -1 \\ 10 & 4 \end{bmatrix}$$

2.

$$\begin{bmatrix} 3 & 25 \\ 5 & 2 \\ 4 & -5 \end{bmatrix}$$

3.

$$\begin{bmatrix} 5 & -1 & 11 \\ -4 & -10 & 2 \\ 2 & 3 & -6 \end{bmatrix}$$

1. $5$

5.

$$\begin{bmatrix} 3 & 9 & 13 \\ 2 & 0 & -4 \\ 10 & 3 & 9 \end{bmatrix}$$

6.

$$\begin{bmatrix} 11 & -4 & 6 \\ 7 & -10 & -6 \\ -1 & 2 & -12 \end{bmatrix}$$

7.

$$\begin{bmatrix} -12 & -6 \\ -9 & -6 \end{bmatrix}$$

8.

$$\begin{bmatrix} 10 & 4 & 2 \\ -21 & 12 & -8 \\ -10 & 2.5 & -30 \end{bmatrix}$$