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In this video we look at the[br]subjective addition and

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subtraction of matrices, and we[br]also look at scalar

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multiplication of matrices. To[br]do that, we're going to need

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some matrices, so here are some[br]matrices that I've already

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prepared, and you'll see we've[br]got four matrices here, and

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they've all got different sizes.[br]So the first thing we need to do

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is just remind ourselves as to[br]how we look at the size of a

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matrix, so we count up the[br]number of rows and the number of

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columns. So this matrix A.

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We say is a two by two matrix[br]because it's got two rows and

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two columns. Matrix B's got[br]three rows and two columns, so

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that's a three by two matrix.

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And we can clearly see that[br]matrix C is a two by three

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matrix, two rows and three[br]columns, and matrix D is a three

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by two matrix with three rows[br]and two columns.

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Now, when it comes to adding and[br]subtracting matrices, we can

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only do it when the two matrices[br]have the same size. That is,

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when they got this both got the[br]same number of rows and the same

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number of columns and two[br]matrices that have the same size

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are said to be compatible, and[br]when they're compatible we can

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add them and subtract them.

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So if we return to our four

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matrices. We see that of these[br]four matrices, the only two that

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are compatible or matrix B and[br]matrix D. They have the same

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size 3 rows and two columns. So[br]what that means is that we can

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find B + D and we can find[br]B -- D and we can find D

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-- B. So we can add and subtract[br]matrices, B&D because they have

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the same size.

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Because they're compatible, we[br]can't add A&B because they have

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different sizes. We can't add[br]C&A because they have different

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sizes. Might be worth noting[br]that where we define the matrix

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C transpose. C transpose. That's[br]where the rows become columns

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would have. Two columns, because[br]each row would turn into a

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column, it would have three[br]rows, so C transpose would be a

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three by two matrix. So C[br]transpose is also compatible

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with B&D. So we can add C[br]transpose to be or today, but we

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can't add C to be or today.

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Once we found two matrices that[br]are compatible that these two

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matrices that have got the same[br]size, then we need to know how

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to actually add them up. So we[br]look at our matrices B&D and see

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how we go through this process.[br]So here's our Matrix B and

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here's our Matrix D and I've[br]written them with a plus sign

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between them and underneath I've[br]written them out again with a

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minus sign. So this is B + D and[br]this is B -- D. So how do we do

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the addition? Well, it's quite

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straightforward. All we do[br]is we were adding we add the

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elements that are in the[br]same position. We call that

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the corresponding position.

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So because the five is in the[br]first row on the 1st column and

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the two is in the first row and[br]the first column, they get added

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together. So we do 5 + 2 which[br]is 7 and that gives us the entry

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in the first row and the first[br]column of our answer.

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We do the same with all the[br]elements, so the minus one is in

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the 2nd row and the first[br]column. So we add that to the

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zero in the 2nd row and the[br]first column. So we do minus 1 +

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0, which gives us minus one and[br]we can continue to do that for

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all six elements of the matrix.[br]So 1 + 4 because the one and

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four are in corresponding[br]positions gives us 5 -- 2 + --

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2. Gives us minus four and that[br]goes up here because it's in the

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first row and the second column[br]first row on the second column

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for three and the one get added[br]to give us four and the nought

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and the minus one get added to[br]give us minus one.

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And so that's how we do matrix[br]addition. So just to recap, we

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have to have two matrices that[br]have the same size and then when

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we have two matrices at the same[br]size we add them by adding the

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elements that are in[br]corresponding positions. And so

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the answer we get is the same[br]size as the two matrices that

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we've added together.

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Now the principles of[br]subtraction are exactly the

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same. We deal with elements that[br]are in the corresponding

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positions, but obviously this[br]time we subtract rather than add

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them. So we do 5 -- 2 to get[br]three, we do minus 1 -- 0 to get

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minus one. We do 1 -- 4 to[br]get minus three. That's done the

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elements in the first column[br]with the elements in the second

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column minus 2 -- -- 2 becomes[br]minus 2 + 2, which is 0.

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3 -- 1 gives us 2[br]and 0 -- -- 1 is 0

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+ 1 which is 1.

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And there's our answer. So[br]when we do B -- D, This is the

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answer. Again, a matrix of the[br]same size as B&D, so that

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illustrates how we do matrix[br]addition and subtraction. We

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have to have two matrices[br]which have the same size in

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order to be compatible. And[br]then what we do is we add or

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subtract the elements that are[br]in the same positions. We call

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corresponding elements.

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Now Matrix obviously has the[br]same size itself, so we can

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always add a matrix to itself,[br]and we're going to do that now

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with the Matrix A. So we're[br]going to add matrix A to

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itself, so into a plus a. So[br]here's Matrix A and what adding

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matrix a onto it. And because[br]it's the same matrix, clearly

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it's not the same size that[br]both 2 by two matrices, so we

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go through the standard[br]procedure.

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When we add elements that are[br]in corresponding positions, so

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the four gets added to the[br]four, which gives us 8, the

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three gets added to the three[br]to give us 6 not getting to

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nought, which gives us nought[br]and minus one gets added to

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minus one. To give this minus[br]2.

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So matrix a + A is this matrix[br]here with entries 860 and minus

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two and we used to writing A[br]plus a in a shorthand form as a

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+ A = 2 A.

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One lot of a there's another lot[br]of a gives us two lots of a.

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So this matrix that we found[br]here, we can refer to as 2A.

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And if we look at the entries in[br]this matrix and compare them

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with the entries of a, we see[br]that each of the entries is just

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twice the entries of a 2 * 4 is[br]eight 2 * 3 or 6 two times

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naughties nought 2 * -- 1 is[br]minus two, and so this process

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illustrates how we do we call[br]scalar multiplication. We take a

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matrix and we multiply it by a[br]number. All that happens is that

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every element inside the matrix.

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Gets multiplied by the number,[br]so in this case the number was

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two and we'll do some examples[br]now, but we use a different

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number. So we've seen how we can[br]do scalar multiplication by

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simply multiplying every element[br]inside our matrix by the number.

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The scalar that we're trying to[br]multiply by. So we'll do a

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couple more examples now, so[br]we're going to workout is going

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to five times the matrix B and[br]I'm going to do 1/2 times the

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matrix D. So all I've done is[br]I've written down what matrix B

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is. I'm going to do five times[br]this matrix. So remember the

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rule for scalar multiplication[br]is the scalar, the number that

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we're trying to multiply by[br]multiplies every entry inside

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the matrix. So we get 5[br]* 5 is 20 five 5 * --

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1 is minus five. 5 * 1 is

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5. 5 * -- 2 is minus[br]ten. 5 * 3 is 15 and 5

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* 0 is 0.

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So this is our answer. This is[br]the matrix 5B or scalar five

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times matrix B. Notice that the[br]matrix we get to that answer has

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the same size, the same order as[br]the matrix we started from, and

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that's fairly obvious. That must[br]be the case because all we do is

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we multiply every element inside[br]the matrix by the scalar

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outside. So we are not creating[br]any new entries in the Matrix

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and we're not losing any. So the[br]matrix that we get must have the

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same size. So when we[br]started with.

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Here's another example. We don't[br]have to just multiply by whole

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numbers or two previous[br]examples. We did 2 * A and 5 *

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B, but we can multiply by any[br]number, and in this case I'm

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choosing to multiply the[br]fraction or half. So I'm going

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to do half times matrix D.

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So here it is written out.[br]Here's matrix D. We do 1/2 times

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that number. All we have to do[br]is do 1/2 times every element

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inside the matrix.

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So we do 1/2 * 2, which gives us[br]one 1/2 * 0, which gives us

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nought 1/2. Times 4, which gives[br]us two. That's not the first

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column and 1/2 * -- 2, giving us[br]minus 1/2 * 1, giving us a half

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and 1/2 * -- 1, giving us minus

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1/2. And so here's our product[br]matrix and not surprisingly, it

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ended up with some fractions.[br]Then, because we were

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multiplying by a fraction to[br]start off with.

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That concludes the video on[br]addition and subtraction of

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matrices and on scalar[br]multiplication.