77.1 Matrix Addition and Subtraction Problem Set

MATRIX ADDITION & SUBTRACTION | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: Given the following matrices, the matrix defined by 𝐴 − 𝐵 is best represented as: 7 3 𝐴=[5 9] 11 −2 10 A. [−3 14 10 B. [ 3 14 10 C. [−3 4 10 D. [11 49

−3 0 𝐵=[ 8 1] −3 −4

3 8] 2 3 8] −2 3 3] 2 7 8] 2

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PROBLEM 2: Given the matrices defined below, the resultant of the 𝐴 + 𝐵, is best written as: 5 2 𝐴=[4 9] 10 −3

−11 0 𝐵=[ 7 1] −6 −8

−6 2 A. [ 11 10 ] 4 −11 0 2 B. [0 10] 4 11 −6 9 C. [ 0 0] 4 1 9 2 D. [1 10] 4 21

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PROBLEM 3: Carrying out the followng operatin results in: 8 [ 0

3 4 5 −2 ]+[ −1 9 6 3

1 ] 5

0 1 3 A. [ ] 6 0 14 −13 −1 4 B. [ ] 6 7 14 1 13 5 C. [ ] 5 2 4 13 1 5 D. [ ] 6 2 14

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MATRIX ADDITION & SUBTRACTION | SOLUTIONS SOLUTION 1: The TOPIC of MATRIX ADDITION & SUBTRACTION can be referenced under the topic of MATRICES on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. ADDITION and SUBTRACTION of two matrices is POSSIBLY ONLY IF both MATRICES have equivalent DIMENSIONS, meaning that they are the same shape and size, ROWS and COLUMNS. The process of ADDING or SUBTRACTING MATRICES simply comes down to taking any of the ELEMENTS in one MATRIX and ADDING of SUBTRACTING that value with the corresponding ELEMENT in another MATRIX. The result is a new ELEMENT that is then placed in to the new MATRIX in the same location of the two elements used in this operation. The GENERAL FORMULA for the ADDITION of TWO MATRICES can be referenced under the topic of MATRICES on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The SUBTRACTION of matrices is represented by the general expression: 𝐴 [ 𝐷

𝐵 𝐸

𝐺 𝐶 ]−[ 𝐽 𝐹

𝐻 𝐾

𝐼 𝐴−𝐺 ]=[ 𝐿 𝐷−𝐽

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𝐵−𝐻 𝐸−𝐾

𝐶−𝐼 ] 𝐹−𝐿

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To illustrate this further, given Matrix “𝐴” and “𝐵” defined as: 𝑎11 𝐴 = [𝑎21 𝑎31

𝑎12 𝑎22 𝑎32

𝑎13 𝑎23 ] 𝑎34

𝑏11 𝐵 = [𝑏21 𝑏31

𝑏12 𝑏22 𝑏32

𝑏13 𝑏23 ] 𝑏34

The addition of “𝐴” and “𝐵” would be: 𝑎11 𝐴 − 𝐵 = [𝑎21 𝑎31

𝑎12 𝑎22 𝑎32

𝑎13 𝑏11 𝑎23 ] − [𝑏21 𝑎34 𝑏31

𝑏12 𝑏22 𝑏32

𝑏13 𝑎11 − 𝑏11 𝑏23 ] = [𝑎21 − 𝑏21 𝑏34 𝑎31 − 𝑏31

𝑎12 − 𝑏12 𝑎22 − 𝑏22 𝑎32 − 𝑏32

𝑎13 − 𝑏13 𝑎23 − 𝑏23 ] 𝑎34 − 𝑏34

In this problem, we are given: 7 3 𝐴=[5 9] 11 −2

−3 0 𝐵=[ 8 1] −3 −4

Through observation, we can confirm that the DIMENSION of MATRIX A is 3x2 and the DIMENSION of MATRIX B is 3x2. The SOLUTION in route to creating the resulting new MATRIX will run along these lines: 𝑎11 𝐴 − 𝐵 = [𝑎21 𝑎31

𝑎12 𝑏11 𝑎22 ] − [𝑏21 𝑎32 𝑏31

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𝑏12 𝑎11 − 𝑏11 𝑏22 ] = [𝑎21 − 𝑏21 𝑏32 𝑎31 − 𝑏31

𝑎12 − 𝑏12 𝑎22 − 𝑏22 ] 𝑎32 − 𝑏32

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Now it’s just a matter of matching up the ELEMENTS and carrying out the ADDITION OPERATION at each location. Doing so we get: 7 𝐴−𝐵 =[5 11

3 −3 0 9 ]−[ 8 1] −3 −4 −2

Let’s walk through and HIGHLIGHT the first few ELEMENTS to illustrate specifically how MATRIX SUBTRACTION is carried out. Starting with ELEMENT 𝑎11 and 𝑏11 , we will focus on the two values: 7 𝐴−𝐵 =[5 11

3 −3 0 9 ]−[ 8 1] −3 −4 −2

Which results in the OPERATION: 7 𝐴−𝐵 =[5 11

7 − (−3) 3 −3 0 9 ]−[ 8 1 ]=[ −3 −4 −2

7 𝐴−𝐵 =[5 11

3 −3 0 10 9 ]−[ 8 1 ]=[ −3 −4 −2

]

Or:

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Now moving on to ELEMENTS 𝑎12 and 𝑏12 , we will focus on the two values: 7 𝐴−𝐵 =[5 11

3 −3 0 10 9 ]−[ 8 1 ]=[ −3 −4 −2

]

Which results in the OPERATION: 7 𝐴−𝐵 =[5 11

3 −3 0 10 3 − 0 ] 9 ]−[ 8 1 ]=[ −3 −4 −2

7 𝐴−𝐵 =[5 11

3 −3 0 10 ] − [ ] = [ 9 8 1 −3 −4 −2

Or: 3 ]

Now filling out the remaining ELEMENT OPERATIONS, we get: 7 𝐴−𝐵 =[5 11

10 3 3 −3 0 9−1 ] 9 ]−[ 8 1 ] = [ 5−8 11 − (−3) −2 − (−4) −3 −4 −2

We conclude that: 7 𝐴−𝐵 =[5 11

3 −3 0 10 3 9 ]−[ 8 1 ] = [−3 8] −3 −4 −2 14 2 Made with

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This is our new MATRIX derived from the SUBTRACTION of MATRIX A and MATRIX B. 𝟏𝟎 The correct answer choice is A. [−𝟑 𝟏𝟒

𝟑 𝟖] 𝟐

SOLUTION 2: The TOPIC of MATRIX ADDITION & SUBTRACTION can be referenced under the topic of MATRICES on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. ADDITION and SUBTRACTION of two matrices is POSSIBLY ONLY IF both MATRICES have equivalent DIMENSIONS, meaning that they are the same shape and size, ROWS and COLUMNS. The process of ADDING or SUBTRACTING MATRICES simply comes down to taking any of the ELEMENTS in one MATRIX and ADDING of SUBTRACTING that value with the corresponding ELEMENT in another MATRIX. The result is a new ELEMENT that is then placed in to the new MATRIX in the same location of the two elements used in this operation. The GENERAL FORMULA for the ADDITION of TWO MATRICES can be referenced under the topic of MATRICES on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with

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The ADDITION of matrices is represented by the general expression: 𝐴 [ 𝐷

𝐵 𝐸

𝐺 𝐶 ]+[ 𝐽 𝐹

𝐻 𝐾

𝐼 𝐴+𝐺 ]=[ 𝐿 𝐷+𝐽

𝐵+𝐻 𝐸+𝐾

𝐶+𝐼 ] 𝐹+𝐿

To illustrate this further, given Matrix “𝐴” and “𝐵” defined as: 𝑎11 𝐴 = [𝑎21 𝑎31

𝑎12 𝑎22 𝑎32

𝑎13 𝑎23 ] 𝑎34

𝑏11 𝐵 = [𝑏21 𝑏31

𝑏12 𝑏22 𝑏32

𝑏13 𝑏23 ] 𝑏34

The addition of “𝐴” and “𝐵” would be: 𝑎11 𝐴 + 𝐵 = [𝑎21 𝑎31

𝑎12 𝑎22 𝑎32

𝑎13 𝑏11 𝑎23 ] + [𝑏21 𝑎34 𝑏31

𝑏12 𝑏22 𝑏32

𝑏13 𝑎11 + 𝑏11 𝑏23 ] = [𝑎21 + 𝑏21 𝑏34 𝑎31 + 𝑏31

𝑎12 + 𝑏12 𝑎22 + 𝑏22 𝑎32 + 𝑏32

𝑎13 + 𝑏13 𝑎23 + 𝑏23 ] 𝑎34 + 𝑏34

In this problem, we are given: 5 𝐴=[4 10

2 9] −3

−11 𝐵=[ 7 −6

0 1] −8

Through observation, we can confirm that the DIMENSION of MATRIX A is 3x2 and the DIMENSION of MATRIX B is 3x2.

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The SOLUTION in route to creating the resulting new MATRIX will run along these lines: 𝑎11 𝐴 + 𝐵 = [𝑎21 𝑎31

𝑎12 𝑏11 𝑎22 ] + [𝑏21 𝑎32 𝑏31

𝑏12 𝑎11 + 𝑏11 𝑏22 ] = [𝑎21 + 𝑏21 𝑏32 𝑎31 + 𝑏31

𝑎12 + 𝑏12 𝑎22 + 𝑏22 ] 𝑎32 + 𝑏32

Now it’s just a matter of matching up the ELEMENTS and carrying out the ADDITION OPERATION at each location. Doing so we get: 5 𝐴+𝐵 =[4 10

−11 0 2 9 ]+[ 7 1] −6 −8 −3

Let’s walk through and HIGHLIGHT the first few ELEMENTS to illustrate specifically how MATRIX ADDITION is carried out. Starting with ELEMENT 𝑎11 and 𝑏11 , we will focus on the two values: 5 𝐴+𝐵 =[4 10

−11 0 2 9 ]+[ 7 1] −6 −8 −3

Which results in the OPERATION: 5 𝐴+𝐵 =[4 10

5 + (−11) −11 0 2 9 ]+[ 7 1 ]=[ −6 −8 −3 Made with

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Or: 5 𝐴+𝐵 =[4 10

−11 0 2 6 9 ]+[ 7 1 ]=[ −6 −8 −3

]

Now moving on to ELEMENTS 𝑎12 and 𝑏12 , we will focus on the two values: 5 𝐴+𝐵 =[4 10

−11 0 2 6 9 ]+[ 7 1 ]=[ −6 −8 −3

]

Which results in the OPERATION: 5 𝐴+𝐵 =[4 10

−11 0 6 2 9 ]+[ 7 1 ]=[ −6 −8 −3

2+0 ]

5 𝐴+𝐵 =[4 10

−11 0 6 2 9 ]+[ 7 1 ]=[ −6 −8 −3

2

Or:

]

Now filling out the remaining ELEMENT OPERATIONS, we get: 5 𝐴+𝐵 =[4 10

6 2 −11 0 2 9+1 ] 9 ]+[ 7 1 ] = [ 4+7 10 + (−6) −3 + (−4) −6 −8 −3

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We conclude that: 5 𝐴+𝐵 =[4 10

−11 0 −6 2 2 9 ]+[ 7 1 ] = [ 11 10 ] −6 −8 −3 4 −11

This is our new MATRIX derived from the ADDITION of MATRIX A and MATRIX B. −𝟔 The correct answer choice is A. [ 𝟏𝟏 𝟒

𝟐 𝟏𝟎 ] −𝟏𝟏

SOLUTION 3: The TOPIC of MATRIX ADDITION & SUBTRACTION can be referenced under the topic of MATRICES on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. ADDITION and SUBTRACTION of two matrices is POSSIBLY ONLY IF both MATRICES have equivalent DIMENSIONS, meaning that they are the same shape and size, ROWS and COLUMNS. The process of ADDING or SUBTRACTING MATRICES simply comes down to taking any of the ELEMENTS in one MATRIX and ADDING of SUBTRACTING that value with the corresponding ELEMENT in another MATRIX. The result is a new ELEMENT that is then placed in to the new MATRIX in the same location of the two elements used in this operation. Made with

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The GENERAL FORMULA for the ADDITION of TWO MATRICES can be referenced under the topic of MATRICES on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The addition of matrices is represented by the general expression: 𝐴 [ 𝐷

𝐵 𝐸

𝐺 𝐶 ]+[ 𝐽 𝐹

𝐻 𝐾

𝐼 𝐴+𝐺 ]=[ 𝐿 𝐷+𝐽

𝐵+𝐻 𝐸+𝐾

𝐶+𝐼 ] 𝐹+𝐿

To illustrate this further, given Matrix “𝐴” and “𝐵” defined as: 𝑎11 𝐴 = [𝑎21 𝑎31

𝑎12 𝑎22 𝑎32

𝑎13 𝑎23 ] 𝑎34

𝑏11 𝐵 = [𝑏21 𝑏31

𝑏12 𝑏22 𝑏32

𝑏13 𝑏23 ] 𝑏34

The addition of “𝐴” and “𝐵” would be: 𝑎11 𝐴 + 𝐵 = [𝑎21 𝑎31

𝑎12 𝑎22 𝑎32

𝑎13 𝑏11 𝑎23 ] + [𝑏21 𝑎34 𝑏31

𝑏12 𝑏22 𝑏32

𝑏13 𝑎11 + 𝑏11 𝑏23 ] = [𝑎21 + 𝑏21 𝑏34 𝑎31 + 𝑏31

𝑎12 + 𝑏12 𝑎22 + 𝑏22 𝑎32 + 𝑏32

𝑎13 + 𝑏13 𝑎23 + 𝑏23 ] 𝑎34 + 𝑏34

In this problem, we are given: 8 3 𝐴=[ 0 −1

4 ] 9

5 −2 𝐵=[ 6 3

1 ] 5

Through observation, we can confirm that the DIMENSION of MATRIX A is 2x3 and the DIMENSION of MATRIX B is 2x3.

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The SOLUTION in route to creating the resulting new MATRIX will run along these lines: 𝑎11 𝐴 + 𝐵 = [𝑎 21

𝑎12 𝑎22

𝑎13 𝑏11 ] + [ 𝑎23 𝑏21

𝑏12 𝑏22

𝑏13 𝑎 + 𝑏11 ] = [ 11 𝑏23 𝑎21 + 𝑏21

𝑎12 + 𝑏12 𝑎22 + 𝑏22

𝑎13 + 𝑏13 ] 𝑎23 + 𝑏23

Now it’s just a matter of matching up the ELEMENTS and carrying out the ADDITION OPERATION at each location. Doing so we get: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] 0 −1 9 6 3 5 Let’s walk through and HIGHLIGHT the first few ELEMENTS to illustrate specifically how MATRIX ADDITION is carried out. Starting with ELEMENT 𝑎11 and 𝑏11 , we will focus on the two values: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] 0 −1 9 6 3 5 Which results in the OPERATION: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] = [8 + 5 0 −1 9 6 3 5

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Or: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] = [13 0 −1 9 6 3 5

]

Now moving on to ELEMENTS 𝑎12 and 𝑏12 , we will focus on the two values: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] = [13 0 −1 9 6 3 5

]

Which results in the OPERATION: 13 3 + (−2) 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ]=[ 0 −1 9 6 3 5

]

Or: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] = [13 0 −1 9 6 3 5

1

]

Now moving on to ELEMENTS 𝑎13 and 𝑏13 , we will focus on the two values: 8 3 4 5 −2 1 𝐴+𝐵 =[ ]+[ ] = [13 0 −1 9 6 3 5

1

]

1

4+1 ]

Which results in the OPERATION: 8 3 4 13 5 −2 1 𝐴+𝐵 =[ ]+[ ]=[ 0 −1 9 6 3 5 Made with

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Or: 8 3 4 5 −2 1 13 𝐴+𝐵 =[ ]+[ ]=[ 0 −1 9 6 3 5

1

5

]

Now filling out the remaining ELEMENT OPERATIONS, we get: 8 3 4 13 5 −2 1 𝐴+𝐵 =[ ]+[ ]=[ 0 −1 9 6 3 5 0+6

1 −1 + 3

5 ] 9+5

And: 8 3 4 5 −2 1 13 1 𝐴+𝐵 =[ ]+[ ]=[ 0 −1 9 6 3 5 6 2

5 ] 14

This is our new MATRIX derived from the ADDITION of MATRIX A and MATRIX B.

The correct answer choice is D. [

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𝟏𝟑 𝟏 𝟓 ] 𝟔 𝟐 𝟏𝟒

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