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This printable supports Common Core Mathematics Standard HSN-VM.C.9

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# Properties of Matrix Multiplication (Grades 11-12)

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## Properties of Matrix Multiplication

1.
Which of the following are always true for matrices $A, B,$ and $C ?$ Choose all that apply.
1. $(AB)C = A(BC)$
2. $A(B+C) = AB + AC$
3. $AB + C = C + BA$
4. $A(B + C) = (B + C)A$
2.
For matrices A, B, C, D, which of the following is always equivalent to the expression $A(B+C)D ?$ There may be more than one correct answer.
1. $A D (B+C)$
2. $ABD + ACD$
3. $(AB +AC)D$
4. $A (BD + CD)$
3.
For the matrices $A = [[6,5],[0,0]]$ and $B = [[0,0],[-4,2]]$, $AB = BA$.
1. True
2. False
4.
For $A = [[-1,0],[0,9]]$ and $B = [[7,0],[0,2]]$, $AB = BA$.
1. True
2. False
5.
For the matrices $A = [[3,4,-1],[0,1,0],[-3,-4,9]], B = [[1,1,1],[1,1,1],[1,1,1]]$, $AB = BA .$
1. True
2. False
6.
Can the following matrix expression be evaluated? Why or why not?

$[[3,-4],[0,-1],[9,1]] \ * ( [[0,1],[1,0]] + [[-5,-9],[-1,4]] )$
1. Yes, but only as is (one cannot distribute the 3-by-2 matrix).
2. Yes, either as is, or if the distributive property is applied.
3. No, after performing the addition of the two smaller matrices, the new dimensions will not allow for multiplication with the left-most matrix.
4. No, all matrices must be have the same dimensions.
7.
For the matrix expression $A (B + C)$, the following is given: $AB = [[7,24],[63,54]], \ \ AC = [[20,-2],[72,18]]$. What is $A (B + C)$ equal to?
1. Not enough information to determine this.
2. $[[27,22],[135,72]]$
3. $[[1868,418],[5148,846]]$
4. $[[13.5, 11],[67.5, 36]]$
8.
Jennie is talking with Liz after their math class. Jennie says, from what she understood from the class, that for matrices $A$ and $B$, $AB$ is never equal to $BA$. Liz thinks that, although this is true in general, there are certain matrices for which $AB = BA$. Who is correct, and why?
1. Jennie is correct. Since the distributive property does not apply, $AB != BA$, for all matrices.
2. Liz is correct. One simply has to take the transpose of the two matrices (the rows become columns of each matrix), and then it is usually true that $AB = BA$.
3. Liz is correct. For example, the identity matrix with any square matrix of the same size.
4. Neither is correct. $AB = BA$ is always true for any matrices A, B.
9.
Simon was given a simple matrix equation, but some of the entries of the first matrix are missing. The equation is as follows.

$([[a,1],[-2,b]] \ [[5,2],[-3,5]] ) \ [[0,-1],[1,0]] = [[13,-17],[-9,7]]$

Simon's work to solve this is as follows:
$L.H.S. = ([[a,1],[-2,b]] \ [[5,2],[-3,5]] ) \ [[0,-1],[1,0]]$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = [[a,1],[-2,b]] \ ([[5,2],[-3,5]] \ [[0,-1],[1,0]] )$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = [[a,1],[-2,b]] \ [[2,-5],[5,3]]$

The first element of the first row will be $2a+5$ and the first element of the second row will be $-4+5b$. Since these have to be equal to the elements of the matrix on the right hand side, this means that $2a+5 = 13$ and $-4+5b = -9$. Solving these equations yields $a = 4$ and $b=-1$.

Is Simon's work correct? If not, where did he make a mistake?
1. This is correct.
2. This is not correct. Simon cannot move the parenthesis as he did in his first step.
3. This is not correct. Simon made a mistake in multiplying the two matrices.
4. This is not correct. Simon only did part of the matrix multiplication to find the two equations. The full matrix multiplication will yield four equations, leading to an overdetermined system of equations that is inconsistent.
10.
In Leane's math homework, she has to multiply three matrices together by hand, as follows: $[[8,2,0],[-4,7,-3]] * [[4,7,-4,-2],[1,3,8,7],[-5,-3,6,6]] * [[-6],[5],[-4],[1]]$. She decides that if she multiplies the last two matrices together first, it will make the computation easier. Is she correct? Why or why not.
1. No, she cannot do this. She has to multiply the first two matrices together first, and then the third, according to normal evaluating rules.
2. Although she is allowed to do this, it will not make the computation easier. Regardless of how she multiplies these matrices, the answer will be a 2-by-1 matrix, and the same work will be involved either way.
3. Yes, she can do this, and it will make the computation easier. Multiplying the second and third matrices together first, and then the first by the product just found, will reduce the total number of arithmetic computations (multiplication and addition) to be performed.
4. These matrices are not able to be multiplied together, as they have incompatible dimensions.
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