79.1 Matrix Multiplication Problem Set
MATRIX MULTIPLICATION | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: Multiply the following matrices: 3 1 6
A.
B.
C.
D.
−1 0 2 4 1
−1 11
2 2
−4 −14 4 8 21 6 4 5 14 4 1 4 4 92 7 1 2 2 1 1 1 4 21 9 9 2 7 9 2 3 9 21 3 9 5 3
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PROBLEM 2: The resulting matrix from the stated operation is most close to: cos (60) sin (60) 0
−sin (60) cos (60) 0
0 2 0 4 1 0
1 A. 2 0 5 B. 3.45 1 −2.46 C. 3.73 0 2.22 D. 7.33 1.11
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PROBLEM 3: Given the matrices below, determine BA: −2 𝐴= 3 0 A. 2
2
52
B. −11
15
C. 23 D. 7
1 −1 2
12 11
7 0 −1
𝐵= 4
−1
5
23 0
2
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PROBLEM 4: Given the matrices below, the operation 𝐴𝐵 is best written as:
𝐴= 2
A. 2
3
5
D. 34
4 −6 2
7 2 1
13
B. 104 C. 0
7
9 𝐵= 3 1
1
2
5
3 4
27
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PROBLEM 5: Given the matrices below, determine AB:
𝐴=
2 7
3 7
1 6
7 9
2 6 𝐵= −2 0
9 3 4 −3
5 3 4 1
−1 2 7 6
20 10 30 53 44 81 89 103 1 0 3 22 B. 5 34 90 −1 −1 −1 5 −5 C. 3 8 0 0 97 44 23 43 D. 12 −1 3 65 A.
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PROBLEM 6: Given the matrices below, determine AB: −2 𝐴= 3 0 A. 7
11
B. −11 C. 23
1 −1 2
7 0 −1
𝐵= 4
−1
5
2 15
12
23 0
D. Not Defined
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MATRIX MULTIPLICATION | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of ROWS in a leading MATRIX with the NUMBER of COLUMNS in a secondary MATRIX. Given the MATRIX A and MATRIX B as: 𝑎!! 𝐴 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝐵=
𝑏!! 𝑏!"
𝑏!" 𝑏!!
𝑏!" 𝑏!"
We can define MATRIX A as a 2x3 MATRIX and MATRIX B as a 3x2 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the
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NUMBER of COLUMNS in the leading MATRIX A must EQUAL the NUMBER of ROWS in the secondary MATRIX B, assuming the OPERATION AB. If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑏 𝑎!! !! 𝑎!" 𝑏!"
𝑏!" 𝑏!!
𝑏!" 𝑏!"
𝑎!! 𝑏!! + 𝑎!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑎!" 𝑏!" + 𝑎!! 𝑏!! 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
In this problem, we are given the OPERATION: 3 𝐴𝐵 = 𝐶 = 1 6
−1 0 2 4 1
−1 11
2 2
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES. By observation, we can see that we have a 3x2 MATRIX and a 2x3 MATRIX. The NUMBER of COLUMNS in MATRIX A is the same as the NUMBER of ROWS in MATRIX B, therefore, MATRIX MULTIPLICATION can be carried out.
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Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑏 𝑎!! !! 𝑎!" 𝑏!"
𝑏!" 𝑏!!
𝑏!" 𝑏!"
𝑎!! 𝑏!! + 𝑎!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑎!" 𝑏!" + 𝑎!! 𝑏!! 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest. Building on our template, we can define the ELEMENTS in MATRIX A as: 3 𝐶= 1 6
−1 𝑏 !! 2 𝑏!" 1
𝑏!" 𝑏!!
𝑏!" 𝑏!"
Doing the same for MATRIX B, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 2
This OPERATION will result in a new 3x3 MATRIX C, such that: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
𝑐!! 2 = 𝑐!" 2 𝑐!"
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𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
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Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
𝑎!! 𝑏!! + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!! and ignoring everything else: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
𝑎!! 𝑏!! + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
We are being asked to MULTIPLY element 𝑎!! and 𝑏!! , which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
𝑎!! 𝑏!! + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling out MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
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3(0) + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
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Let’s cycle right to the next portion in defining this ELEMENT, which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
3(0) + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling in MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
3(0) + −1 4 𝑐!" 𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!! , we have: 3(0) + −1 4 𝑐!" 𝑐!"
3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Or: 𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Let’s move to defining ELEMENT 𝑐!" , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
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Let’s highlight each step in defining this specific ELEMENT, starting with the term 𝑎!! 𝑏!" and ignoring everything else: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
We are being asked to MULTIPLY element 𝑎!! and 𝑏!" , which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
3(−1) + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling out MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
Let’s cycle right to the next portion in defining this ELEMENT, which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
3(−1) + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling in MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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3(−1) + −1 11 𝑐!! 𝑐!"
𝑐!! 𝑐!" 𝑐!!
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Highlighting the full operation for defining ELEMENT 𝑐!" , we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
3(−1) + −1 11 𝑐!! 𝑐!"
3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Or: 𝑐!" 𝑐!" 𝑐!!
Moving over to ELEMENT 𝑐!" , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!" and ignoring everything else: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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−14 𝑐!! 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
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We are being asked to MULTIPLY element 𝑎!! and 𝑏!" , which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
−14 𝑐!! 𝑐!"
3(2) + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
And filling out MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
Let’s cycle right to the next portion in defining this ELEMENT, which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
3(2) + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
−14 𝑐!! 𝑐!"
3(2) + (−1)(2) 𝑐!" 𝑐!!
And filling in MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!" , we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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−14 𝑐!! 𝑐!"
3(2) + (−1)(2) 𝑐!" 𝑐!!
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Or: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
4 𝑐!" 𝑐!!
So I think you see by now the various OPERATIONS will continue on allowing us to define each ELEMENT within the resulting MATRIX C. This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENT to MUTLIPLY against. Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 1 0 +2 4 2 6 0 +1 4
3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
−14 1 −1 + 2(11) 6 −1 + 1 11
4 1 2 +2 2 6 2 +1 2
And: −4 8 4
−14 21 5
−𝟒 The correct answer choice is A. 𝟖 𝟒
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4 6 14
−𝟏𝟒 𝟐𝟏 𝟓
𝟒 𝟔 𝟏𝟒
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SOLUTION 2: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of ROWS in a leading MATRIX with the NUMBER of COLUMNS in a secondary MATRIX. Given the MATRIX A and MATRIX B as: 𝑎!! 𝐴 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝐵 = 𝑏!" 𝑏!"
We can define MATRIX A as a 3x3 MATRIX and MATRIX B as a 3x1 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in the leading MATRIX A must EQUAL the NUMBER of ROWS in secondary MATRIX B.
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If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!" + 𝑎!! 𝑏!"
In this problem, we are given the OPERATION: cos (60) 𝐶 = sin (60) 0
−sin (60) cos (60) 0
0 2 0 4 1 0
Converting the trigonometric expressions in to more user friendly terms: .5 𝐶 = . 866 0
−.866 .5 0
0 2 0 4 1 0
We now must confirm that we can actually MULTIPLY these two MATRICES. By observation, we can see that we have a 3x3 MATRIX and a 3x1 MATRIX. The NUMBER of COLUMNS in the leading MATRIX is the same as the NUMBER of ROWS in the secondary MATRIX, therefore, MATRIX MULTIPLICATION can be carried out.
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Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!" + 𝑎!! 𝑏!"
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest. Building on our template, we can define the ELEMENTS in MATRIX A as: .5 𝐶 = . 866 0
−.866 .5 0
0 𝑏!! 0 𝑏!" 1 𝑏!"
Doing the same for MATRIX B, we have: .5 𝐶 = . 866 0
−.866 .5 0
0 2 0 4 1 0
This OPERATION will result in a new 3x1 MATRIX C, such that: .5 𝐶 = . 866 0
−.866 .5 0
𝑐!! 0 2 0 4 = 𝑐!" 𝑐!" 1 0
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Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: .5 𝐶 = . 866 0
−.866 .5 0
𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!! and ignoring everything else: .5 𝐶 = . 866 0
−.866 .5 0
𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
We are being asked to MULTIPLY element 𝑎!! and 𝑏!! , which is: .5 𝐶 = . 866 0
−.866 .5 0
𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
And filling out MATRIX C, we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Let’s cycle right to the next portion in defining this ELEMENT, which is: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
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And filling in MATRIX C, we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Cycling once more to the right: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Filling in MATRIX C, we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 0(0) 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Highlighting the full operation for defining ELEMENT 𝑐!! , we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 0(0) 0 2 𝑐!" 0 4 = 𝑐!" 1 0
.5 𝐶 = . 866 0
−.866 .5 0
−2.46 0 2 0 4 = 𝑐!" 𝑐!" 1 0
Or:
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You can now see how each OPERATION will play out in defining each ELEMENT within the resulting MATRIX C. This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because a ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENTS to MUTLIPLY against. Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us: .5 𝐶 = . 866 0
−.866 .5 0
−2.46 0 2 0 4 = . 866 2 + .5 4 + 0(0) 0 2 + 0 4 + 1(0) 1 0
.5 𝐶 = . 866 0
−.866 .5 0
−2.46 0 2 0 4 = 3.73 0 1 0
Or:
−𝟐. 𝟒𝟔 The correct answer choice is C. 𝟑. 𝟕𝟑 𝟎
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SOLUTION 3: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of ROWS in a leading MATRIX with the NUMBER of COLUMNS in a secondary MATRIX. Given the MATRIX A and MATRIX B as: 𝑎!! 𝑎!" 𝑎!" 𝐴 = 𝑎!" 𝑎!! 𝑎!" 𝐵 = 𝑏!! 𝑎!" 𝑎!" 𝑎!!
𝑏!"
𝑏!"
We can define MATRIX A as a 3x3 MATRIX and MATRIX B as a 1x3 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in the leading MATRIX must EQUAL the NUMBER of ROWS in the secondary MATRIX.
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If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as:
𝐶 = 𝑏!!
𝑏!"
𝑏!"
𝑎!! 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
= 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!" + 𝑏!" 𝑎!! + 𝑏!" 𝑎!"
𝑏!! 𝑎!" + 𝑏!" 𝑎!" + 𝑏!" 𝑎!!
In this problem, we are given the OPERATION:
𝐵𝐴 = 𝐶 = 4
−1
−2 5 3 0
1 −1 2
7 0 −1
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES in the order that is requested. By observation, we can see that we have a 1x3 MATRIX and a 3x3 MATRIX. The NUMBER of COLUMNS in MATRIX B is the same as the NUMBER of ROWS in MATRIX A, therefore, MATRIX MULTIPLICATION can be carried out. However, if the ORDER of MULTIPLICATION was reversed, meaning we were asked to assess AB, the OPERATION could not be carried out due to the fact that we wouldn’t have a scenario where the NUMBER of COLUMNS in the leading MATRIX matched the NUMBER of ROWS in the secondary MATRIX.
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Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have:
𝐶 = 𝑏!!
𝑏!"
𝑏!"
𝑎!! 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
= 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!" + 𝑏!" 𝑎!! + 𝑏!" 𝑎!"
𝑏!! 𝑎!" + 𝑏!" 𝑎!" + 𝑏!" 𝑎!!
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest. Building on our template, we can define the ELEMENTS in MATRIX B as:
𝐶= 4
−1
𝑎!! 5 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
Doing the same for MATRIX A, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 −1
This OPERATION will result in a new 1x3 MATRIX C, such that:
𝐶= 4
−1
−2 5 3 0
1 −1 2
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7 0 = 𝑐!! −1
𝑐!"
𝑐!"
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Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑏!! 𝑎!! and ignoring everything else:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
𝑐!"
𝑐!"
𝑐!"
𝑐!"
We are being asked to MULTIPLY element 𝑏!! and 𝑎!! , which is:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
And filling out MATRIX C, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
Let’s cycle right to the next portion in defining this ELEMENT, which is:
𝐶= 4
−1
−2 5 3 0
1 −1 2
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𝑐!"
𝑐!"
And filling in MATRIX C, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + −1 3 + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
7 0 = 4(−2) + −1 3 + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
Cycling once more to the right:
𝐶= 4
−1
−2 5 3 0
1 −1 2
Filling in MATRIX C, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + −1 3 + 5(0) −1
𝑐!"
𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!! , we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + −1 3 + 5(0) −1
−1
−2 5 3 0
1 −1 2
7 0 = −11 −1
𝑐!"
𝑐!"
Or: 𝐶= 4
𝑐!"
𝑐!"
You can now see how each OPERATION will continue as we define each ELEMENT within the resulting MATRIX C.
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This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENTS to MUTLIPLY against. Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = −8 − 3 + 0 −1
−1
−2 5 3 0
1 −1 2
7 0 = −11 −1
4 + 1 + 10
28 + 0 − 5
Or:
𝐶= 4
The correct answer choice is B. −𝟏𝟏
𝟏𝟓
15
23
𝟐𝟑
SOLUTION 4: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of COLUMNS in a leading MATRIX with the NUMBER of ROWS in a secondary MATRIX. Given the MATRIX A and MATRIX B as:
𝐴 = 𝑎!!
𝑎!"
𝑏!! 𝐵 = 𝑏!" 𝑏!"
𝑎!"
𝑏!" 𝑏!! 𝑏!"
𝑏!" 𝑏!" 𝑏!!
We can define MATRIX A as a 1x3 MATRIX and MATRIX B as a 3x3 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in MATRIX A must EQUAL the NUMBER of ROWS in MATRIX B. If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as:
𝐶 = 𝑎!!
𝑎!"
𝑎!"
𝑏!! 𝑏!" 𝑏!"
𝑏!" 𝑏!! 𝑏!"
= 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
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𝑏!" 𝑏!" 𝑏!!
𝑎!! 𝑏!" + 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
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In this problem, we are given the OPERATION:
𝐴𝐵 = 𝐶 = 2
3
7
9 3 1
4 −6 2
7 2 1
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES in the order that is requested. By observation, we can see that we have a 1x3 MATRIX and a 3x3 MATRIX. The NUMBER of COLUMNS in MATRIX A is the same as the NUMBER of ROWS in MATRIX B, therefore, MATRIX MULTIPLICATION can be carried out. However, if the ORDER of MULTIPLICATION was reversed, meaning we were asked to assess BA, the OPERATION could not be carried out due to the fact that we wouldn’t have a scenario where the NUMBER of COLUMNS in the leading MATRIX matched the NUMBER of ROWS in the secondary MATRIX. Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have:
𝐶 = 𝑎!!
𝑎!"
𝑎!"
𝑏!! 𝑏!" 𝑏!"
𝑏!" 𝑏!! 𝑏!"
= 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
𝑏!" 𝑏!" 𝑏!!
𝑎!! 𝑏!! + 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest.
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Building on our template, we can define the ELEMENTS in MATRIX A as:
𝐶= 2
3
𝑏!! 7 𝑏!" 𝑏!"
𝑏!" 𝑏!! 𝑏!"
𝑏!" 𝑏!" 𝑏!!
Doing the same for MATRIX B, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 1
This OPERATION will result in a new 1x3 MATRIX C, such that:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑐!! 1
𝑐!"
𝑐!"
Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!! and ignoring everything else:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
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𝑐!"
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𝑐!"
We are being asked to MULTIPLY element 𝑎!! and 𝑏!! , which is:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
And filling out MATRIX C, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
Let’s cycle right to the next portion in defining this ELEMENT, which is:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
And filling in MATRIX C, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
𝑐!"
𝑐!"
Cycling once more to the right:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 𝑎!" 𝑏!" 1
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Filling in MATRIX C, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 7(1) 1
𝑐!"
𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!! , we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 7(1) 1
3
9 7 3 1
4 −6 2
7 2 = 34 1
𝑐!"
𝑐!"
Or:
𝐶= 2
𝑐!"
𝑐!"
You can now see how each ELEMENT will be defined within the resulting MATRIX C. This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENT to MUTLIPLY against.
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Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 34 1
8 − 18 + 14
3
9 7 3 1
4 −6 2
7 2 = 34 1
4
14 + 6 + 7
Or:
𝐶= 2
The correct answer choice is D. 𝟑𝟒
27
𝟒
𝟐𝟕
SOLUTION 5: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. Given the MATRIX A and MATRIX B as:
𝐴=
2 7
3 7
2 9 5 −1 1 7 6 3 3 2 𝐵= 6 9 −2 4 4 7 0 −3 1 6 Made with by Prepineer | Prepineer.com
We can define the DEMINSION of our MATRIX A as a 2x4 MATRIX and MATRIX B as a 4x4 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in MATRIX A must EQUAL the NUMBER of ROWS in MATRIX B. Since we have 4 COLUMNS in MATRIX A and 4 ROWS in MATRIX B, our RELATIONSHIP is confirmed and we can move forward with the OPERATION. In this problem, we are asked to determine:
𝐴𝐵 =
2 7
3 7
1 6
2 7 6 9 −2 0
9 3 4 −3
5 3 4 1
−1 2 7 6
It is important to note that ORDER MATTERS in this particular OPERATION. If the MULTIPLICATION was reversed, meaning we were asked to assess BA, the OPERATION could not be carried out due to the fact that we wouldn’t have a scenario where the NUMBER of COLUMNS in the leading MATRIX matched the NUMBER of ROWS in the secondary MATRIX.
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Bypassing the creation of each individual ELEMENT, the RESULTANT MATRIX C from the product of AB, can be written as:
𝐶=
2 7
3 7
1 6
2 7 6 9 −2 0
9 3 4 −3
5 3 4 1
The correct answer choice is A.
−1 2 20 = 7 44 6 𝟐𝟎 𝟒𝟒
𝟏𝟎 𝟖𝟏
𝟑𝟎 𝟖𝟗
10 81
30 89
53 103
𝟓𝟑 𝟏𝟎𝟑
SOLUTION 6: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of COLUMNS in a leading MATRIX with the NUMBER of ROWS in a secondary MATRIX.
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Given the MATRIX A and MATRIX B as: 𝑎!! 𝐴 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝐵 = 𝑏!!
𝑏!"
𝑏!"
We can define MATRIX A as a 3x3 MATRIX and MATRIX B as a 1x3 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in MATRIX A must EQUAL the NUMBER of ROWS in MATRIX B. In this problem, we are given the OPERATION: −2 𝐴𝐵 = 3 0
1 −1 2
7 0 4 −1
−1
5
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES in the order that is requested. By observation, we can see that we have a 3x3 MATRIX and a 1x3 MATRIX. The NUMBER of COLUMNS in the leading MATRIX A does not match the NUMBER of ROWS in MATRIX B, therefore, MATRIX MULTIPLICATION can’t be carried out. The correct answer choice is D. Not Defined
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PROBLEM 1: Multiply the following matrices: 3 1 6
A.
B.
C.
D.
−1 0 2 4 1
−1 11
2 2
−4 −14 4 8 21 6 4 5 14 4 1 4 4 92 7 1 2 2 1 1 1 4 21 9 9 2 7 9 2 3 9 21 3 9 5 3
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PROBLEM 2: The resulting matrix from the stated operation is most close to: cos (60) sin (60) 0
−sin (60) cos (60) 0
0 2 0 4 1 0
1 A. 2 0 5 B. 3.45 1 −2.46 C. 3.73 0 2.22 D. 7.33 1.11
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PROBLEM 3: Given the matrices below, determine BA: −2 𝐴= 3 0 A. 2
2
52
B. −11
15
C. 23 D. 7
1 −1 2
12 11
7 0 −1
𝐵= 4
−1
5
23 0
2
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PROBLEM 4: Given the matrices below, the operation 𝐴𝐵 is best written as:
𝐴= 2
A. 2
3
5
D. 34
4 −6 2
7 2 1
13
B. 104 C. 0
7
9 𝐵= 3 1
1
2
5
3 4
27
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PROBLEM 5: Given the matrices below, determine AB:
𝐴=
2 7
3 7
1 6
7 9
2 6 𝐵= −2 0
9 3 4 −3
5 3 4 1
−1 2 7 6
20 10 30 53 44 81 89 103 1 0 3 22 B. 5 34 90 −1 −1 −1 5 −5 C. 3 8 0 0 97 44 23 43 D. 12 −1 3 65 A.
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PROBLEM 6: Given the matrices below, determine AB: −2 𝐴= 3 0 A. 7
11
B. −11 C. 23
1 −1 2
7 0 −1
𝐵= 4
−1
5
2 15
12
23 0
D. Not Defined
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MATRIX MULTIPLICATION | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of ROWS in a leading MATRIX with the NUMBER of COLUMNS in a secondary MATRIX. Given the MATRIX A and MATRIX B as: 𝑎!! 𝐴 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝐵=
𝑏!! 𝑏!"
𝑏!" 𝑏!!
𝑏!" 𝑏!"
We can define MATRIX A as a 2x3 MATRIX and MATRIX B as a 3x2 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the
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NUMBER of COLUMNS in the leading MATRIX A must EQUAL the NUMBER of ROWS in the secondary MATRIX B, assuming the OPERATION AB. If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑏 𝑎!! !! 𝑎!" 𝑏!"
𝑏!" 𝑏!!
𝑏!" 𝑏!"
𝑎!! 𝑏!! + 𝑎!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑎!" 𝑏!" + 𝑎!! 𝑏!! 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
In this problem, we are given the OPERATION: 3 𝐴𝐵 = 𝐶 = 1 6
−1 0 2 4 1
−1 11
2 2
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES. By observation, we can see that we have a 3x2 MATRIX and a 2x3 MATRIX. The NUMBER of COLUMNS in MATRIX A is the same as the NUMBER of ROWS in MATRIX B, therefore, MATRIX MULTIPLICATION can be carried out.
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Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑏 𝑎!! !! 𝑎!" 𝑏!"
𝑏!" 𝑏!!
𝑏!" 𝑏!"
𝑎!! 𝑏!! + 𝑎!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑎!" 𝑏!" + 𝑎!! 𝑏!! 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest. Building on our template, we can define the ELEMENTS in MATRIX A as: 3 𝐶= 1 6
−1 𝑏 !! 2 𝑏!" 1
𝑏!" 𝑏!!
𝑏!" 𝑏!"
Doing the same for MATRIX B, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 2
This OPERATION will result in a new 3x3 MATRIX C, such that: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
𝑐!! 2 = 𝑐!" 2 𝑐!"
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𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
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Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
𝑎!! 𝑏!! + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!! and ignoring everything else: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
𝑎!! 𝑏!! + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
We are being asked to MULTIPLY element 𝑎!! and 𝑏!! , which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
𝑎!! 𝑏!! + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling out MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
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3(0) + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
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Let’s cycle right to the next portion in defining this ELEMENT, which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
3(0) + 𝑎!" 𝑏!" 𝑐!" 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling in MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
3(0) + −1 4 𝑐!" 𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!! , we have: 3(0) + −1 4 𝑐!" 𝑐!"
3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Or: 𝑐!" 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Let’s move to defining ELEMENT 𝑐!" , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
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Let’s highlight each step in defining this specific ELEMENT, starting with the term 𝑎!! 𝑏!" and ignoring everything else: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
We are being asked to MULTIPLY element 𝑎!! and 𝑏!" , which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
3(−1) + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling out MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
Let’s cycle right to the next portion in defining this ELEMENT, which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
3(−1) + 𝑎!" 𝑏!! 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
And filling in MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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3(−1) + −1 11 𝑐!! 𝑐!"
𝑐!! 𝑐!" 𝑐!!
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Highlighting the full operation for defining ELEMENT 𝑐!" , we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
3(−1) + −1 11 𝑐!! 𝑐!"
3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
𝑐!" 𝑐!" 𝑐!!
Or: 𝑐!" 𝑐!" 𝑐!!
Moving over to ELEMENT 𝑐!" , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!" and ignoring everything else: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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−14 𝑐!! 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
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We are being asked to MULTIPLY element 𝑎!! and 𝑏!" , which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
−14 𝑐!! 𝑐!"
3(2) + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
And filling out MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
Let’s cycle right to the next portion in defining this ELEMENT, which is: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
3(2) + 𝑎!" 𝑏!" 𝑐!" 𝑐!!
−14 𝑐!! 𝑐!"
3(2) + (−1)(2) 𝑐!" 𝑐!!
And filling in MATRIX C, we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!" , we have: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
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−14 𝑐!! 𝑐!"
3(2) + (−1)(2) 𝑐!" 𝑐!!
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Or: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 𝑐!" 2 𝑐!"
−14 𝑐!! 𝑐!"
4 𝑐!" 𝑐!!
So I think you see by now the various OPERATIONS will continue on allowing us to define each ELEMENT within the resulting MATRIX C. This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENT to MUTLIPLY against. Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us: 3 𝐶= 1 6
−1 0 2 4 1
−1 11
−4 2 = 1 0 +2 4 2 6 0 +1 4
3 𝐶= 1 6
−1 0 2 4 1
−1 11
2 = 2
−14 1 −1 + 2(11) 6 −1 + 1 11
4 1 2 +2 2 6 2 +1 2
And: −4 8 4
−14 21 5
−𝟒 The correct answer choice is A. 𝟖 𝟒
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4 6 14
−𝟏𝟒 𝟐𝟏 𝟓
𝟒 𝟔 𝟏𝟒
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SOLUTION 2: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of ROWS in a leading MATRIX with the NUMBER of COLUMNS in a secondary MATRIX. Given the MATRIX A and MATRIX B as: 𝑎!! 𝐴 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝐵 = 𝑏!" 𝑏!"
We can define MATRIX A as a 3x3 MATRIX and MATRIX B as a 3x1 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in the leading MATRIX A must EQUAL the NUMBER of ROWS in secondary MATRIX B.
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If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!" + 𝑎!! 𝑏!"
In this problem, we are given the OPERATION: cos (60) 𝐶 = sin (60) 0
−sin (60) cos (60) 0
0 2 0 4 1 0
Converting the trigonometric expressions in to more user friendly terms: .5 𝐶 = . 866 0
−.866 .5 0
0 2 0 4 1 0
We now must confirm that we can actually MULTIPLY these two MATRICES. By observation, we can see that we have a 3x3 MATRIX and a 3x1 MATRIX. The NUMBER of COLUMNS in the leading MATRIX is the same as the NUMBER of ROWS in the secondary MATRIX, therefore, MATRIX MULTIPLICATION can be carried out.
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Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have: 𝑎!! 𝐶 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 𝑏!" = 𝑎!" 𝑏!! + 𝑎!! 𝑏!" + 𝑎!" 𝑏!" 𝑏!" 𝑎!" 𝑏!! + 𝑎!" 𝑏!" + 𝑎!! 𝑏!"
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest. Building on our template, we can define the ELEMENTS in MATRIX A as: .5 𝐶 = . 866 0
−.866 .5 0
0 𝑏!! 0 𝑏!" 1 𝑏!"
Doing the same for MATRIX B, we have: .5 𝐶 = . 866 0
−.866 .5 0
0 2 0 4 1 0
This OPERATION will result in a new 3x1 MATRIX C, such that: .5 𝐶 = . 866 0
−.866 .5 0
𝑐!! 0 2 0 4 = 𝑐!" 𝑐!" 1 0
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Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION: .5 𝐶 = . 866 0
−.866 .5 0
𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!! and ignoring everything else: .5 𝐶 = . 866 0
−.866 .5 0
𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
We are being asked to MULTIPLY element 𝑎!! and 𝑏!! , which is: .5 𝐶 = . 866 0
−.866 .5 0
𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
And filling out MATRIX C, we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Let’s cycle right to the next portion in defining this ELEMENT, which is: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
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And filling in MATRIX C, we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Cycling once more to the right: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 𝑎!" 𝑏!" 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Filling in MATRIX C, we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 0(0) 0 2 𝑐!" 0 4 = 𝑐!" 1 0
Highlighting the full operation for defining ELEMENT 𝑐!! , we have: .5 𝐶 = . 866 0
−.866 .5 0
. 5(2) + (−.866)4 + 0(0) 0 2 𝑐!" 0 4 = 𝑐!" 1 0
.5 𝐶 = . 866 0
−.866 .5 0
−2.46 0 2 0 4 = 𝑐!" 𝑐!" 1 0
Or:
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You can now see how each OPERATION will play out in defining each ELEMENT within the resulting MATRIX C. This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because a ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENTS to MUTLIPLY against. Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us: .5 𝐶 = . 866 0
−.866 .5 0
−2.46 0 2 0 4 = . 866 2 + .5 4 + 0(0) 0 2 + 0 4 + 1(0) 1 0
.5 𝐶 = . 866 0
−.866 .5 0
−2.46 0 2 0 4 = 3.73 0 1 0
Or:
−𝟐. 𝟒𝟔 The correct answer choice is C. 𝟑. 𝟕𝟑 𝟎
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SOLUTION 3: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of ROWS in a leading MATRIX with the NUMBER of COLUMNS in a secondary MATRIX. Given the MATRIX A and MATRIX B as: 𝑎!! 𝑎!" 𝑎!" 𝐴 = 𝑎!" 𝑎!! 𝑎!" 𝐵 = 𝑏!! 𝑎!" 𝑎!" 𝑎!!
𝑏!"
𝑏!"
We can define MATRIX A as a 3x3 MATRIX and MATRIX B as a 1x3 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in the leading MATRIX must EQUAL the NUMBER of ROWS in the secondary MATRIX.
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If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as:
𝐶 = 𝑏!!
𝑏!"
𝑏!"
𝑎!! 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
= 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!" + 𝑏!" 𝑎!! + 𝑏!" 𝑎!"
𝑏!! 𝑎!" + 𝑏!" 𝑎!" + 𝑏!" 𝑎!!
In this problem, we are given the OPERATION:
𝐵𝐴 = 𝐶 = 4
−1
−2 5 3 0
1 −1 2
7 0 −1
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES in the order that is requested. By observation, we can see that we have a 1x3 MATRIX and a 3x3 MATRIX. The NUMBER of COLUMNS in MATRIX B is the same as the NUMBER of ROWS in MATRIX A, therefore, MATRIX MULTIPLICATION can be carried out. However, if the ORDER of MULTIPLICATION was reversed, meaning we were asked to assess AB, the OPERATION could not be carried out due to the fact that we wouldn’t have a scenario where the NUMBER of COLUMNS in the leading MATRIX matched the NUMBER of ROWS in the secondary MATRIX.
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Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have:
𝐶 = 𝑏!!
𝑏!"
𝑏!"
𝑎!! 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
= 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝑏!! 𝑎!" + 𝑏!" 𝑎!! + 𝑏!" 𝑎!"
𝑏!! 𝑎!" + 𝑏!" 𝑎!" + 𝑏!" 𝑎!!
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest. Building on our template, we can define the ELEMENTS in MATRIX B as:
𝐶= 4
−1
𝑎!! 5 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
Doing the same for MATRIX A, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 −1
This OPERATION will result in a new 1x3 MATRIX C, such that:
𝐶= 4
−1
−2 5 3 0
1 −1 2
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7 0 = 𝑐!! −1
𝑐!"
𝑐!"
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Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑏!! 𝑎!! and ignoring everything else:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
𝑐!"
𝑐!"
𝑐!"
𝑐!"
We are being asked to MULTIPLY element 𝑏!! and 𝑎!! , which is:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 𝑏!! 𝑎!! + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
And filling out MATRIX C, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + 𝑏!" 𝑎!" + 𝑏!" 𝑎!" −1
Let’s cycle right to the next portion in defining this ELEMENT, which is:
𝐶= 4
−1
−2 5 3 0
1 −1 2
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𝑐!"
𝑐!"
And filling in MATRIX C, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + −1 3 + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
7 0 = 4(−2) + −1 3 + 𝑏!" 𝑎!" −1
𝑐!"
𝑐!"
Cycling once more to the right:
𝐶= 4
−1
−2 5 3 0
1 −1 2
Filling in MATRIX C, we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + −1 3 + 5(0) −1
𝑐!"
𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!! , we have:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = 4(−2) + −1 3 + 5(0) −1
−1
−2 5 3 0
1 −1 2
7 0 = −11 −1
𝑐!"
𝑐!"
Or: 𝐶= 4
𝑐!"
𝑐!"
You can now see how each OPERATION will continue as we define each ELEMENT within the resulting MATRIX C.
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This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENTS to MUTLIPLY against. Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us:
𝐶= 4
−1
−2 5 3 0
1 −1 2
7 0 = −8 − 3 + 0 −1
−1
−2 5 3 0
1 −1 2
7 0 = −11 −1
4 + 1 + 10
28 + 0 − 5
Or:
𝐶= 4
The correct answer choice is B. −𝟏𝟏
𝟏𝟓
15
23
𝟐𝟑
SOLUTION 4: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.
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Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of COLUMNS in a leading MATRIX with the NUMBER of ROWS in a secondary MATRIX. Given the MATRIX A and MATRIX B as:
𝐴 = 𝑎!!
𝑎!"
𝑏!! 𝐵 = 𝑏!" 𝑏!"
𝑎!"
𝑏!" 𝑏!! 𝑏!"
𝑏!" 𝑏!" 𝑏!!
We can define MATRIX A as a 1x3 MATRIX and MATRIX B as a 3x3 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in MATRIX A must EQUAL the NUMBER of ROWS in MATRIX B. If this is the case, then we can move forward with calculating the product of MATRIX A and MATRIX B using a SYSTEMATIC approach illustrated as:
𝐶 = 𝑎!!
𝑎!"
𝑎!"
𝑏!! 𝑏!" 𝑏!"
𝑏!" 𝑏!! 𝑏!"
= 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
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𝑏!" 𝑏!" 𝑏!!
𝑎!! 𝑏!" + 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
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In this problem, we are given the OPERATION:
𝐴𝐵 = 𝐶 = 2
3
7
9 3 1
4 −6 2
7 2 1
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES in the order that is requested. By observation, we can see that we have a 1x3 MATRIX and a 3x3 MATRIX. The NUMBER of COLUMNS in MATRIX A is the same as the NUMBER of ROWS in MATRIX B, therefore, MATRIX MULTIPLICATION can be carried out. However, if the ORDER of MULTIPLICATION was reversed, meaning we were asked to assess BA, the OPERATION could not be carried out due to the fact that we wouldn’t have a scenario where the NUMBER of COLUMNS in the leading MATRIX matched the NUMBER of ROWS in the secondary MATRIX. Revisiting the SYSTEMATIC process of MATRIX MULTIPLICATION, we have:
𝐶 = 𝑎!!
𝑎!"
𝑎!"
𝑏!! 𝑏!" 𝑏!"
𝑏!" 𝑏!! 𝑏!"
= 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!"
𝑏!" 𝑏!" 𝑏!!
𝑎!! 𝑏!! + 𝑎!" 𝑏!! + 𝑎!" 𝑏!"
𝑎!! 𝑏!" + 𝑎!" 𝑏!" + 𝑎!" 𝑏!!
We will lay this process out step by step, defining a few elements from scratch, and then fill in the rest.
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Building on our template, we can define the ELEMENTS in MATRIX A as:
𝐶= 2
3
𝑏!! 7 𝑏!" 𝑏!"
𝑏!" 𝑏!! 𝑏!"
𝑏!" 𝑏!" 𝑏!!
Doing the same for MATRIX B, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 1
This OPERATION will result in a new 1x3 MATRIX C, such that:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑐!! 1
𝑐!"
𝑐!"
Let’s start off with defining ELEMENT 𝑐!! , in our resultant MATRIX C. We will proceed with defining this ELEMENT using the OPERATION:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
Let’s highlight each step in defining this specific ELEMENT starting with the term 𝑎!! 𝑏!! and ignoring everything else:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
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𝑐!"
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𝑐!"
We are being asked to MULTIPLY element 𝑎!! and 𝑏!! , which is:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 𝑎!! 𝑏!! + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
And filling out MATRIX C, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
Let’s cycle right to the next portion in defining this ELEMENT, which is:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 𝑎!" 𝑏!" + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
And filling in MATRIX C, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 𝑎!" 𝑏!" 1
𝑐!"
𝑐!"
𝑐!"
𝑐!"
Cycling once more to the right:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 𝑎!" 𝑏!" 1
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Filling in MATRIX C, we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 7(1) 1
𝑐!"
𝑐!"
Highlighting the full operation for defining ELEMENT 𝑐!! , we have:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 2(9) + 3(3) + 7(1) 1
3
9 7 3 1
4 −6 2
7 2 = 34 1
𝑐!"
𝑐!"
Or:
𝐶= 2
𝑐!"
𝑐!"
You can now see how each ELEMENT will be defined within the resulting MATRIX C. This also highlights why it’s MANDATORY that the two matrices being MULTIPLIED share that COLUMN to ROW relationship. If they didn’t, the OPERATION would fail because ELEMENTS in one MATRIX will eventually not have any corresponding ELEMENT to MUTLIPLY against.
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Bypassing the combination of terms, filling in the remainder of the resultant MATRIX C gives us:
𝐶= 2
3
9 7 3 1
4 −6 2
7 2 = 34 1
8 − 18 + 14
3
9 7 3 1
4 −6 2
7 2 = 34 1
4
14 + 6 + 7
Or:
𝐶= 2
The correct answer choice is D. 𝟑𝟒
27
𝟒
𝟐𝟕
SOLUTION 5: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. Given the MATRIX A and MATRIX B as:
𝐴=
2 7
3 7
2 9 5 −1 1 7 6 3 3 2 𝐵= 6 9 −2 4 4 7 0 −3 1 6 Made with by Prepineer | Prepineer.com
We can define the DEMINSION of our MATRIX A as a 2x4 MATRIX and MATRIX B as a 4x4 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in MATRIX A must EQUAL the NUMBER of ROWS in MATRIX B. Since we have 4 COLUMNS in MATRIX A and 4 ROWS in MATRIX B, our RELATIONSHIP is confirmed and we can move forward with the OPERATION. In this problem, we are asked to determine:
𝐴𝐵 =
2 7
3 7
1 6
2 7 6 9 −2 0
9 3 4 −3
5 3 4 1
−1 2 7 6
It is important to note that ORDER MATTERS in this particular OPERATION. If the MULTIPLICATION was reversed, meaning we were asked to assess BA, the OPERATION could not be carried out due to the fact that we wouldn’t have a scenario where the NUMBER of COLUMNS in the leading MATRIX matched the NUMBER of ROWS in the secondary MATRIX.
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Bypassing the creation of each individual ELEMENT, the RESULTANT MATRIX C from the product of AB, can be written as:
𝐶=
2 7
3 7
1 6
2 7 6 9 −2 0
9 3 4 −3
5 3 4 1
The correct answer choice is A.
−1 2 20 = 7 44 6 𝟐𝟎 𝟒𝟒
𝟏𝟎 𝟖𝟏
𝟑𝟎 𝟖𝟗
10 81
30 89
53 103
𝟓𝟑 𝟏𝟎𝟑
SOLUTION 6: The GENERAL FORMULA for MULTIPLICATION of two MATRICES can be referenced under the topic of MATHEMATICS on page 34 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Given a pair of matrices, we are able to MULTIPLY the ELEMENTS from each of the matrices using a specific SYSTEMATIC approach, defining each ELEMENT in a new unique MATRIX in the process. In MATRIX MULTIPLICATION, we need a set of MATRICES that share the NUMBER of COLUMNS in a leading MATRIX with the NUMBER of ROWS in a secondary MATRIX.
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Given the MATRIX A and MATRIX B as: 𝑎!! 𝐴 = 𝑎!" 𝑎!"
𝑎!" 𝑎!! 𝑎!"
𝑎!" 𝑎!" 𝑎!!
𝐵 = 𝑏!!
𝑏!"
𝑏!"
We can define MATRIX A as a 3x3 MATRIX and MATRIX B as a 1x3 MATRIX. In order for MULTIPLICATION to be possible on these two defined MATRICES, then we need to confirm a COLUMN to ROW relationship between them…meaning the NUMBER of COLUMNS in MATRIX A must EQUAL the NUMBER of ROWS in MATRIX B. In this problem, we are given the OPERATION: −2 𝐴𝐵 = 3 0
1 −1 2
7 0 4 −1
−1
5
The first thing we must do is confirm that we can actually MULTIPLY these two MATRICES in the order that is requested. By observation, we can see that we have a 3x3 MATRIX and a 1x3 MATRIX. The NUMBER of COLUMNS in the leading MATRIX A does not match the NUMBER of ROWS in MATRIX B, therefore, MATRIX MULTIPLICATION can’t be carried out. The correct answer choice is D. Not Defined
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