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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (13)

1 answer

Last reply by: Dr Carleen Eaton
Thu Jul 31, 2014 7:15 PM

Post by Philippe Tremblay on July 28, 2014

At the very end, in your review, you state that you multiplied (B+C) by A. Beware as this is not the same as A(B+C). As you explained earlier, the commutative property does not apply to matrices.

2 answers

Last reply by: Dr Carleen Eaton
Tue Feb 19, 2013 8:25 PM

Post by Carroll Fields on February 17, 2013

ar approx. 28:30 into lecture, the calculation of example IV row 2 col 1, should be '0' not '1' as listed above. Making the final answer [5 13]
[0 -3]

similar problem as the one listed by Jeff Mitchell, please correct to prevent misinformation from being given.

1 answer

Last reply by: Carroll Fields
Sun Feb 17, 2013 12:50 PM

Post by julius mogyorossy on March 15, 2012

I like to imagine turning the first column 90 degrees to the right to see if they are multicompatible, then I keep turning the rows in the first column 90 degrees to the right, works for me.

5 answers

Last reply by: Dr Carleen Eaton
Thu Jul 31, 2014 7:13 PM

Post by Jeff Mitchell on March 6, 2011

at approx. 9:45 into lecture, the calculation for the value in row 2 column 1 --> the math was incorrectly done for 2(-2) which was simplified to (-2) and should have been -4. therefore, the sum value for Row 2,column 1 should be 4 and not 6.

Matrix Multiplication

  • Matrix multiplication is defined if and only if the number of columns of the first matrix is equal to the number of rows of the second matrix.
  • Matrix multiplication is associative and satisfies the left and right distributive properties.
  • Matrix multiplication is not commutative.

Matrix Multiplication

Find the product of matrix A and B
A = [
2
3
4
5
]B = [
1
2
4
5
]
  • Since the demensions of matrix A and matrix B match, you can proceed with multiplication.
  • One intuitive way to multiply matrices is to put the second matrix on top of the second one like this:




  • 1
    2
    4
    5






    2
    3
    4
    5









  • The result from Rows and Columns will fall right into place on the resulting matrix.




  • 1
    2
    4
    5






    2
    3
    4
    5






    R1C1
    R1C2
    R2C1
    R2C2



    =



    1
    2
    4
    5






    2
    3
    4
    5






    R1C1
    R1C2
    R2C1
    R2C2



    =



    1
    2
    4
    5






    2
    3
    4
    5






    (2)(1) + (3)(4)
    (2)(2) + (3)(5)
    (4)(1) + (5)(4)
    (4)(2) + (5)(5)



  • Simplify the resulting matrix
  • [
    (2)(1) + (3)(4)
    (2)(2) + (3)(5)
    (4)(1) + (5)(4)
    (4)(2) + (5)(5)
    ] = [
    2 + 12
    4 + 15
    4 + 20
    8 + 25
    ] = [
    14
    19
    24
    33
    ]




14
19
24
33



Find the product of matrix A and BA = [
− 1
2
3
− 2
]B = [
1
− 2
− 1
0
]
  • Since the demensions of matrix A and matrix B match, you can proceed with multiplication.
  • One intuitive way to multiply matrices is to put the second matrix on top of the second one like this:




  • 1
    − 2
    − 1
    0






    − 1
    2
    3
    − 2









  • The result from Rows and Columns will fall right into place on the resulting matrix.



    1
    − 2
    − 1
    0






    − 1
    2
    3
    − 2






    R1C1
    R1C2
    R2C1
    R2C2



    =



    1
    − 2
    − 1
    0






    − 1
    2
    3
    − 2






    R1C1
    R1C2
    R2C1
    R2C2



    =



    1
    − 2
    − 1
    0






    − 1
    2
    3
    − 2






    ( − 1)(1) + (2)( − 1)
    ( − 1)( − 2) + (2)(0)
    (3)(1) + ( − 2)( − 1)
    (3)( − 2) + ( − 2)(0)



  • Simplify the resulting matrix




  • ( − 1)(1) + (2)( − 1)
    ( − 1)( − 2) + (2)(0)
    (3)(1) + ( − 2)( − 1)
    (3)( − 2) + ( − 2)(0)



    =


    − 1 + − 2
    2 + 0
    3 + 2
    − 6 + 0



    =


    − 3
    2
    5
    − 6







− 3
2
5
− 6



Find the product of matrix A and BA = [
− 1
2
3
− 2
]B = [
1
0
0
1
]
  • Since the demensions of matrix A and matrix B match, you can proceed with multiplication.
  • One intuitive way to multiply matrices is to put the second matrix on top of the second one like this:




  • 1
    0
    0
    1






    − 1
    2
    3
    − 2









  • The result from Rows and Columns will fall right into place on the resulting matrix.




  • 1
    0
    0
    1






    − 1
    2
    3
    − 2






    R1C1
    R1C2
    R2C1
    R2C2



    =



    1
    0
    0
    1






    − 1
    2
    3
    − 2






    R1C1
    R1C2
    R2C1
    R2C2



    =



    1
    0
    0
    1






    − 1
    2
    3
    − 2






    ( − 1)(1) + (2)(0)
    ( − 1)(0) + (2)(1)
    (3)(1) + ( − 2)(0)
    (3)(0) + ( − 2)(1)



  • Simplify the resulting matrix




  • ( − 1)(1) + (2)(0)
    ( − 1)(0) + (2)(1)
    (3)(1) + ( − 2)(0)
    (3)(0) + ( − 2)(1)



    =


    − 1 + 0
    0 + 2
    3 + 0
    0 − 2



    =


    − 1
    2
    3
    − 2







− 1
2
3
− 2



Find the product of matrix A and BA = [
1
− 2
− 1
0
]B = [
− 1
2
3
− 2
]
  • Since the demensions of matrix A and matrix B match, you can proceed with multiplication.
  • One intuitive way to multiply matrices is to put the second matrix on top of the second one like this:




  • − 1
    2
    3
    − 2






    1
    − 2
    − 1
    0









  • The result from Rows and Columns will fall right into place on the resulting matrix.




  • − 1
    2
    3
    − 2






    1
    − 2
    − 1
    0






    R1C1
    R1C2
    R2C1
    R2C2



    =



    − 1
    2
    3
    − 2






    1
    − 2
    − 1
    0






    R1C1
    R1C2
    R2C1
    R2C2



    =



    − 1
    2
    3
    − 2






    1
    − 2
    − 1
    0






    (1)( − 1) + ( − 2)(3)
    (1)(2) + ( − 2)( − 2)
    ( − 1)( − 1) + (0)(3)
    ( − 1)(2) + (0)( − 2)



  • Simplify the resulting matrix




  • (1)( − 1) + ( − 2)(3)
    (1)(2) + ( − 2)( − 2)
    ( − 1)( − 1) + (0)(3)
    ( − 1)(2) + (0)( − 2)



    =


    − 1 + − 6
    2 + 4
    1 + 0
    − 2 + 0



    =


    − 7
    6
    1
    − 2







− 7
6
1
− 2



In problem number 4 you multiplied [
1
− 2
− 1
0
]*[
− 1
2
3
− 2
] = [
− 7
6
1
− 2
] and in problem 2 you multiplied [
− 1
2
3
− 2
]*[
1
− 2
− 1
0
] = [
− 3
2
5
− 6
]
  • What do you notice about matrix property of multiplication?
I noticed that matrix multiplication is not commutative. In other words A*B ≠ B*A
Find [
− 2
3
0
0
3
− 1
]*[
− 1
2
1
4
− 2
0
]
  • Since the dimensions match 2x3 and 3x2, 3 = 3. You can proceed with the multiplication.
  • Use the same strategy as with previous problems.





  • − 1
    2
    1
    4
    − 2
    0







    − 2
    3
    0
    0
    3
    − 1














  • − 1
    2
    1
    4
    − 2
    0







    − 2
    3
    0
    0
    3
    − 1






    − 2( − 1) + 3(1) + 0( − 2)
    − 2(2) + 3(4) + 0(0)
    0( − 1) + 3(1) + − 1( − 2)
    0(2) + 3(4) + − 1(0)



  • Simplify the resulting matrix




  • − 2( − 1) + 3(1) + 0( − 2)
    − 2(2) + 3(4) + 0(0)
    0( − 1) + 3(1) + − 1( − 2)
    0(2) + 3(4) + − 1(0)



    =


    2 + 3 + 0
    − 4 + 12 + 0
    0 + 3 + 2
    0 + 12 + 0



    =


    5
    8
    5
    12







5
8
5
12



Find [
− 1
1
− 2
2
4
0
]*[
− 1
2
1
4
− 2
0
]
  • Since the dimensions match 2x3 and 3x2, 3 = 3. You can proceed with the multiplication.
  • Use the same strategy as with previous problems.





  • − 1
    2
    1
    4
    − 2
    0







    − 1
    1
    − 2
    2
    4
    0














  • − 1
    2
    1
    4
    − 2
    0







    − 1
    1
    − 2
    2
    4
    0






    − 1( − 1) + 1(1) + − 2( − 2)
    − 1(2) + 1(4) + − 2(0)
    2( − 1) + 4(1) + 0( − 2)
    2(2) + 4(4) + 0(0)



  • Simplify the resulting matrix




  • − 1( − 1) + 1(1) + − 2( − 2)
    − 1(2) + 1(4) + − 2(0)
    2( − 1) + 4(1) + 0( − 2)
    2(2) + 4(4) + 0(0)



    =


    1 + 1 + 4
    − 2 + 4 + 0
    − 2 + 4 + 0
    4 + 16 + 0



    =


    6
    2
    2
    20







6
2
2
20



Find [
1
0
− 1
1
0
− 1
]*[
− 1
1
0
0
2
− 2
]
  • Since the dimensions match 2x3 and 3x2, 3 = 3. You can proceed with the multiplication.
  • Use the same strategy as with previous problems.





  • − 1
    1
    0
    0
    2
    − 2







    1
    0
    − 1
    1
    0
    − 1














  • − 1
    1
    0
    0
    2
    − 2







    1
    0
    − 1
    1
    0
    − 1






    1( − 1) + 0(0) + − 1(2)
    1(1) + 0(0) + − 1( − 2)
    1( − 1) + 0(0) + − 1(2)
    1(1) + 0(0) + − 1( − 2)



  • Simplify the resulting matrix




  • 1( − 1) + 0(0) + − 1(2)
    1(1) + 0(0) + − 1( − 2)
    1( − 1) + 0(0) + − 1(2)
    1(1) + 0(0) + − 1( − 2)



    =


    − 1 + 0 + − 2
    1 + 0 + 2
    − 1 + 0 + − 2
    1 + 0 + 2



    =


    − 3
    3
    − 3
    3







− 3
3
− 3
3



*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Matrix Multiplication

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Dimension Requirement 0:17
    • n = p
    • Resulting Product Matrix (m x q)
    • Example: Multiplication
  • Matrix Multiplication 3:38
    • Example: Matrix Multiplication
  • Properties of Matrix Multiplication 10:46
    • Associative Property
    • Associative Property (Scalar)
    • Distributive Property
    • Distributive Property (Scalar)
  • Example 1: Possible Matrices 13:31
  • Example 2: Multiplying Matrices 17:08
  • Example 3: Multiplying Matrices 20:41
  • Example 4: Matrix Properties 24:41

Transcription: Matrix Multiplication

Welcome to Educator.com.0000

In today's lesson, we will be covering matrix multiplication.0002

In a previous lesson, we discussed scalar multiplication, which is multiplying a constant by a matrix.0005

This time, we will be talking about multiplying one matrix by another.0012

Before you proceed with matrix multiplication, you need to verify that the dimension requirement has been met.0018

So, suppose that matrix A is m by n, and matrix B has dimensions of p by q.0024

The product of these two matrices can be obtained only if n = p.0031

Now, what is that saying? What that is saying is that the number of columns of the first matrix must equal the number of rows of the second matrix.0036

So, in order to find AB, to find that product, to be allowed to do that kind of multiplication,0045

the number of columns of the first matrix must equal the number of rows of the second matrix.0052

OK, once you find that product, the resulting product matrix, AB, will have dimensions of m times q.0099

So, the product matrix dimensions will have the number of rows of the first matrix and the same number of columns as the second matrix.0109

To make this more concrete, let's look at an example.0115

If matrix A equals 2, -1, 4, 0, and matrix B equals 3, 4, 2, 1, 0, -3, 0, -2; my first question is, "Can I multiply them--can I find AB?"0118

Well, this has one row and four columns; it is a 1x4 matrix.0137

This has four rows and two columns; it is a 4x2 matrix; so this gives me 1x4 and 4x2.0145

So, all I have to do is see, "OK, does this second number equal this first number?" and yes, the second number does equal the first number.0152

Therefore, I can multiply those.0162

Now, what are going to be the dimensions of the product matrix?0164

Well, the dimensions of the product matrix...this is a 1x4, multiplied by a 4x2 matrix; and AB is going to have0168

the same number of rows as this first matrix (which is 1), and the same number of columns as the second matrix (which is 2).0175

So, the product matrix is going to be a 1x2 matrix.0184

So, the important thing is: before you multiply, make sure that you verify that the dimension requirement is met--0188

that the second number of the first matrix is equal to the first number of the second matrix.0193

And to predict the dimensions of the product matrix, you take this first number and this second number; and that will give you the product matrix dimensions, 1x2.0200

OK, multiplying by matrices is not exactly what you would expect.0217

It is not like addition and subtraction of matrices, where (in addition and subtraction) we just took the first matrix0224

and added the corresponding element of that to the corresponding element of the second matrix (and the same in subtraction).0231

You might think, "OK, I am just going to multiply the corresponding elements by each other."0238

But it is actually more complicated than that; and you need to take it step-by-step.0243

So, the best way to understand this is to go through an example: so let's look at a pair of matrices and multiply them out.0247

OK, this is my first matrix; I am going to call it A; and then, I want to multiply it by another matrix, which I am going to call B.0264

So, before I multiply, I have to make sure that they meet the dimension requirements.0277

And A has two rows, and it has four columns; B has four rows and two columns.0284

So, I am allowed to multiply these, because this second number equals the first number.0292

My product matrix, AB, is going to have two rows and two columns; it is going to be a square matrix with dimensions 2x2.0299

Now, as you read up here, it says that the element in row I and column J of the product...0311

so the element in a certain row and column of the product of the matrices A and B...0317

is obtained by forming the sum of the products of the corresponding elements in row I0323

(in a certain row of matrix A) and the corresponding column, J, of matrix B.0329

What does this mean? Well, instead of just saying row I and column 1, let's start with row 1, column 1.0335

I want to find the element that goes right here, in row 1, column 1.0345

And the way I am going to do that is: I am going to go over here to row 1, and I am going to go here in column 1;0350

and I am going to multiply 3 by 4 and find that product, and then I am going to add that to the next product, 2 by 0.0361

Then, I am going to add that to the next product, 0 by 1, to 0 by 3, and so on.0369

So, row 1, column 1; then I am going to go to the corresponding row in A and the corresponding column in B.0375

And working this out, this gives me 3 times 4, plus 2 times 0, plus 0 times 3, plus 1 times -2.0380

Working that out, that is going to give me 12 + 0 + 0 - 2; 12 - 2 is 10.0399

Therefore, my row 1, column 1 element is 10.0410

OK, next let's work on row 1, column 2; that means I am going to go to row 1 here in A, and column 2 here, in B, and do the same operations.0414

So, row 1, column 2: that is 3 times -1, plus 2 times 2, plus 0 times 1, plus 1 times 0.0429

OK, working that out, this gives me -3 + 4 + 0 + 0; that is 4 and -3, so that is 1.0446

Therefore, my row 1, column 2 element for the product is 1.0461

All right, now I need to work with this second row; and I am going to think about what position this is right here.0468

This is row 2, column 1; that means I am going to go to row 2 here and column 1 in B and do the same thing.0472

OK, row 2, column 1: -1 times 4, plus 0 times 0, plus 4 times 3 (so the third element here and the third element here),0487

plus the fourth element here and the fourth element there.0507

All right, figuring this out, this is going to give me -4 + 0 + 12 + -2, or - 2; look at it either way.0512

This is going to give me -4 and 12, which is 8, minus 2...actually, let's make this clearer...plus 12, minus 2...it is going to give me...0526

let's see...that is -4, 0, 4 and 3, 2 and -2; OK, so this is going to give me 12 - 6, or 6.0546

Now, the next row and column position: I have row 2, column 2.0562

Here is row 2; here is column 2; it is going to give me -1 times -1, plus 0 times 2, plus 4 times 1, plus 2 times 0.0577

OK, so that is going to come out to...-1 times -1 is 1, plus 0, plus 4, plus 0; 4 + 1 is 5, so I am going to get 5 right here.0601

So again, if I just looked here, and I said, "OK, that is row 1, column 2," then I would get that0618

by going to row 1 here and column 2 here and multiplying each of those, and then finding the sum of their products.0626

So, even with a bigger matrix, you can always find a particular position by using this method.0638

OK, there are some properties that govern matrix multiplication that you need to be familiar with.0646

If A, B, and C are matrices for which products are defined, and k is any scalar, then these properties hold.0653

The first property you will recognize as the associative property for multiplication.0660

And what it says is that I can either multiply A times B, find that product, and then multiply by C;0667

or I can multiply B times C, find that product, and then multiply A; and I will get the same result.0679

OK, next I see the same thing here (it is the associative property), except this time, it also involves multiplying a scalar.0687

So, here I have matrix multiplication, and then I also have a scalar.0695

And I can either multiply the two matrices and then multiply the scalar times that product;0698

or I can multiply the scalar times the first matrix, then multiply that product by the second matrix;0704

or the scalar and the second matrix, and the product by the first matrix.0712

So, it doesn't matter which order I do the multiplication in, in this situation; these two first, then the other two.0716

Next, you are going to recognize the distributive property.0726

And the distributive property, as usual, says that, if I have these two matrices, B + C,0732

and I am multiplying A by them, another approach that would give the equal result0739

is to first multiply A and B, and then add that product to the product of A and C--the distributive property.0743

And again, the distributive property would work if I placed these as follows, added A and B, and then multiplied C.0751

The same thing: I could say the product of A and C, plus the product of B and C.0763

What is very important to realize is that the order that you multiply matrices in matters a lot.0768

So, if I say, "Oh, I am going to do AB; does this equal BA?" no, it does not always equal that.0775

Sometimes it does; but very often it does not--these are two different things.0783

Therefore, matrix multiplication is not commutative; it does not follow the commutative property,0788

and so you can't simply say, "A times B is the same as B times A"; you can't just change the order of those two.0797

The first example: suppose A is 3x4; B is a matrix that is 4x3; C is 3x3; and D is 4x4.0812

What are possible defined products of the four matrices?0821

Recall that, for a product to be defined, the second number of the first matrix has to equal the first number of the other matrix.0824

So, let's first look at AA; is that defined?0834

Well, if I have 3x4, and I am trying to multiply it by 3x4; these two are not equal, so this is not defined.0840

OK, and let's make a list here of the ones that are defined, because that is what they are asking me for--where the product is defined.0852

If I take A times B, well, the second number of A is equal to the first number of B; so that is defined.0863

Now, let's do it in the opposite order, BA: if I have 4x3 and 3x4, that is also defined.0871

OK, now AC: 3x4 and 3x3--that is not defined.0882

Now, if I put C first--if I say CA--it is 3x3 and 3x4, and these two are equal, so CA would have a defined product.0888

OK, now 3x4 and 4x4; that is AD; that is defined, because of the 4 and the 4.0903

However, if I put D first--if I say I am going to do DA--that is going to give me the D first, which is 4x4, and then A, 3x4; that is not defined.0910

OK, moving on to B: B times B (BB) is 4x3 and 4x3; that is not defined.0923

BC is 4x3 and 3x3; these two are equal, so BC is defined.0934

CB is 3x3 and 4x3--not defined; OK, BD--4x3 and 4x4--not defined.0940

But if I put the D first, then I would get 4x4 and 4x3; that is defined, so DB is defined.0953

OK, now I am up to C; 3x3 and 3x3--CC--that is a square matrix, so that is defined; the multiplication is defined for that.0965

OK, now, CD--3x3 and 4x4--is not defined; then I try the D first--it is not going to matter: 4x4 and 3x3 is still not defined.0975

OK, let's go to D: I can multiply DD, because I have 4x4 and 4x4, and that means that, of course,0988

since this is a square matrix, this second number and the first number there are going to be equal.1000

So, these are my defined products; I have 3, 6, 8 defined products.1005

So, these are all the ones where the second number of the first matrix--1010

the number of columns--was equal to the first number of the second matrix--the number of rows.1016

And I can perform multiplication of these sets of matrices.1021

OK, now doing some matrix multiplication: before I proceed, I am going to make sure that the product is defined--that I can do it.1029

So, I have a 2x2 matrix here, and I have a 2x2 matrix here; therefore, multiplication is allowed.1038

I am rewriting this down here, since this second number is equal to this first number (so multiplication is allowed).1052

Using the method we discussed: my product matrix is going to be 2x2 also--the first number here and the second number there.1066

I am expecting a 2x2 matrix; so let's first look for this position--row 1, column 1.1076

I am going to go to row 1 here and column 1 here and work with those two.1086

I am going to say 3 times 0 (that product), and I am going to add it to the product of 2 times 4.1091

That is going to give me 0 + 8, which is 8; so the element in row 1, column 1 is 8.1099

Now, row 1, column 2--I am going to go to row 1 here--row 1 of this first matrix--and column 2 of the second matrix.1111

And that is going to give me...row 1 is 3, times 6; so row 1, column 2 is 2 and -1, so plus 2 times -1.1124

It is going to give me 18 - 2, which equals 16.1140

OK, now the next position is row 2, column 1; working with that, that is going to give me a -1 times 0, plus 4 times 4,1150

which is going to equal 0 plus 16, which is 16.1170

Next, I want to find...this is row 2, but it is column 2 this time.1177

So, I go to row 2 here and column 2 here: -1 times 6, plus 4 times -1--that is going to give me -6 - 4, or -10.1185

As expected, my product matrix is also 2x2; and again, if I picked any element in here--1208

let's say I picked this 16--I could simply say, "OK, that is row 2, column 1."1215

And I would get that by multiplying and finding the products of row 2 of the first matrix1223

and column 1 of the second matrix, and adding those up--finding the sum of those.1230

This is the result of the multiplication of these two matrices.1236

Example 3: I have 1, 2 rows and 3 columns, so I have a 2x3 matrix and a 3x2 matrix.1244

These two are equal, so I am allowed to multiply them.1255

The product of these two is going to have this number of rows (2) and this number of columns.1258

So, I am going to get a matrix that is going to be 2x2 for my product here.1267

OK, I am setting up the product matrix right here, and first looking for row1 , column 1; so I am going to use row 1 here, column 1 here.1275

2 times 0 is the product; then I am going to add that to the next product, which is -1 times 3, and the third product, which is 0 times -2,1293

which equals...2 times 0 gives me 0, minus 3, plus 0; so it is -3--row 1, column 1, right here, is -3.1314

Row 1 right here, but column 2: I am going to take 2 times 1; OK, in row 1, column 2, the next set of elements is -1 and 6,1331

so plus -1 times 6; then I still have row 1, column 2; I have 0 times -1; figure that out--1355

that is 2 - 6 + 0, so it is just 2 - 6, and that gives me -4, so row 1, column 2 is -4.1368

OK, moving on to the second row: the first thing I need to find is row 2, column 1--row 2 here, column 1 here.1380

0 times 0, plus 3 times 3, plus -2 times -2; OK, that is going to give me 0 + 9, and then -2 times -2 is positive 4, so plus 4.1392

9 + 4 is 13, so row 2, column 1, gives me 13.1412

OK, next I need to find row 2, column 2.1417

Row 2, column 2: 0 times 1, plus 3 times 6, plus -2 times -1 is going to give me 0 + 18... -2 and -1 is positive 2, so that is 20.1425

OK, and as expected, I got a 2x2 matrix, and I could find any position by just honing in and saying,1448

"OK, that is row 1, column 2, so I am going to multiply row 1 elements and column 2 elements and add those products."1454

Now, at first, I know this seems like a lot of work; but you should go step-by-step, find the row, find the column,1462

write out what you need to write out; and then, later on, as you get faster, you can just do a lot of it in your head.1467

But right now, since there are a lot of steps, go to each row and each column, and figure out what those are.1473

OK, Example 4 asks us to do some multiplication and some addition, so I need to find the sum of the products AB + AC.1479

But remembering the distributive property, I can actually do this more easily.1494

Recall that, according to the distributive property for matrix multiplication, AB + AC = A(B + C).1499

I would rather do it this way, because matrix multiplication is difficult; addition is much easier.1516

This way, instead of multiplying twice and adding once, I would rather just add these together, and then only have to multiply one time.1522

So, I am going to approach this by using the distributive property.1532

So, what I am going to do is find A times (B + C), and it is going to be equivalent to this.1536

Let's start out by finding B + C: B is 0, 3, 1, 4; that is B; and I am going to add that to C, which is right here: 1, 2, 0, -3.1543

Now, recall from matrix addition: all you have to do is add the corresponding elements,1567

and you are going to get a matrix of the same dimensions as the original.1572

And these are both 2x2 matrices, so I can add them.1575

0 + 1 is going to give me 1; 3 + 2 is going to give me 5; 1 + 0--I am going to get 1; 4 - 3 is 1.1578

Now, this gives me B + C; the next thing I need to do is multiply A times B + C.1590

So, let's go back up here and look at what A is: A is...2 and -1 are the elements, and 3, 2; that is A.1602

I am going to go ahead and add that to what I discovered that B + C is: 1, 5, 1...oh, excuse me, multiply--we are now multiplying that.1611

All right, recall from matrix multiplication that I have to make sure I am even allowed to multiply these.1624

And this is a 2x2 matrix, and this is a 2x2 matrix; so this second number is equal to the first number here, so I can multiply them.1635

All right, row 1, column 1--this position on my product matrix--is going to give me row 1 here and column 1 here.1644

That is 2 times 1, plus 3 times 1; that is just 2 + 3, so that is 5.1653

OK, now, row 1, column 2: row 1 here, and column 2 here: 2 times 5, plus 3 times 1.1664

2 times 5 is 10, plus 3--that is going to give me 13 for this position.1678

OK, the next row, row 2, column 1: -1 times 1, plus 2 times 1; that is going to give me -1 plus 2, which equals 1.1685

So, for row 2, column 1, I get 1.1707

Now, row 2, column 2, right here, is -1 times 5, plus 2 times 1.1710

-1 times 5 is -5, plus 2 gives me -3.1725

So, this gives me A times (B + C), which is equivalent to what they asked me to find, AB + AC.1732

So, the key step here was recognizing that you could use the distributive property there, because that made a lot less work.1739

I only had to do one set of matrix multiplication, instead of 2.1745

So, I first used the distributive property; and then I had to add B and C.1750

So, I added B and C, and got this sum matrix, B + C.1757

And then, I multiplied it by A; so here is (B + C) times A, using my typical method to get this result.1761

That concludes this lesson of Educator.com on matrix multiplication, and I will see you next time!1770