In this video we look at the subjective addition and subtraction of matrices, and we also look at scalar multiplication of matrices. To do that, we're going to need some matrices, so here are some matrices that I've already prepared, and you'll see we've got four matrices here, and they've all got different sizes. So the first thing we need to do is just remind ourselves as to how we look at the size of a matrix, so we count up the number of rows and the number of columns. So this matrix A. We say is a two by two matrix because it's got two rows and two columns. Matrix B's got three rows and two columns, so that's a three by two matrix. And we can clearly see that matrix C is a two by three matrix, two rows and three columns, and matrix D is a three by two matrix with three rows and two columns. Now, when it comes to adding and subtracting matrices, we can only do it when the two matrices have the same size. That is, when they got this both got the same number of rows and the same number of columns and two matrices that have the same size are said to be compatible, and when they're compatible we can add them and subtract them. So if we return to our four matrices. We see that of these four matrices, the only two that are compatible or matrix B and matrix D. They have the same size 3 rows and two columns. So what that means is that we can find B + D and we can find B -- D and we can find D -- B. So we can add and subtract matrices, B&D because they have the same size. Because they're compatible, we can't add A&B because they have different sizes. We can't add C&A because they have different sizes. Might be worth noting that where we define the matrix C transpose. C transpose. That's where the rows become columns would have. Two columns, because each row would turn into a column, it would have three rows, so C transpose would be a three by two matrix. So C transpose is also compatible with B&D. So we can add C transpose to be or today, but we can't add C to be or today. Once we found two matrices that are compatible that these two matrices that have got the same size, then we need to know how to actually add them up. So we look at our matrices B&D and see how we go through this process. So here's our Matrix B and here's our Matrix D and I've written them with a plus sign between them and underneath I've written them out again with a minus sign. So this is B + D and this is B -- D. So how do we do the addition? Well, it's quite straightforward. All we do is we were adding we add the elements that are in the same position. We call that the corresponding position. So because the five is in the first row on the 1st column and the two is in the first row and the first column, they get added together. So we do 5 + 2 which is 7 and that gives us the entry in the first row and the first column of our answer. We do the same with all the elements, so the minus one is in the 2nd row and the first column. So we add that to the zero in the 2nd row and the first column. So we do minus 1 + 0, which gives us minus one and we can continue to do that for all six elements of the matrix. So 1 + 4 because the one and four are in corresponding positions gives us 5 -- 2 + -- 2. Gives us minus four and that goes up here because it's in the first row and the second column first row on the second column for three and the one get added to give us four and the nought and the minus one get added to give us minus one. And so that's how we do matrix addition. So just to recap, we have to have two matrices that have the same size and then when we have two matrices at the same size we add them by adding the elements that are in corresponding positions. And so the answer we get is the same size as the two matrices that we've added together. Now the principles of subtraction are exactly the same. We deal with elements that are in the corresponding positions, but obviously this time we subtract rather than add them. So we do 5 -- 2 to get three, we do minus 1 -- 0 to get minus one. We do 1 -- 4 to get minus three. That's done the elements in the first column with the elements in the second column minus 2 -- -- 2 becomes minus 2 + 2, which is 0. 3 -- 1 gives us 2 and 0 -- -- 1 is 0 + 1 which is 1. And there's our answer. So when we do B -- D, This is the answer. Again, a matrix of the same size as B&D, so that illustrates how we do matrix addition and subtraction. We have to have two matrices which have the same size in order to be compatible. And then what we do is we add or subtract the elements that are in the same positions. We call corresponding elements. Now Matrix obviously has the same size itself, so we can always add a matrix to itself, and we're going to do that now with the Matrix A. So we're going to add matrix A to itself, so into a plus a. So here's Matrix A and what adding matrix a onto it. And because it's the same matrix, clearly it's not the same size that both 2 by two matrices, so we go through the standard procedure. When we add elements that are in corresponding positions, so the four gets added to the four, which gives us 8, the three gets added to the three to give us 6 not getting to nought, which gives us nought and minus one gets added to minus one. To give this minus 2. So matrix a + A is this matrix here with entries 860 and minus two and we used to writing A plus a in a shorthand form as a + A = 2 A. One lot of a there's another lot of a gives us two lots of a. So this matrix that we found here, we can refer to as 2A. And if we look at the entries in this matrix and compare them with the entries of a, we see that each of the entries is just twice the entries of a 2 * 4 is eight 2 * 3 or 6 two times naughties nought 2 * -- 1 is minus two, and so this process illustrates how we do we call scalar multiplication. We take a matrix and we multiply it by a number. All that happens is that every element inside the matrix. Gets multiplied by the number, so in this case the number was two and we'll do some examples now, but we use a different number. So we've seen how we can do scalar multiplication by simply multiplying every element inside our matrix by the number. The scalar that we're trying to multiply by. So we'll do a couple more examples now, so we're going to workout is going to five times the matrix B and I'm going to do 1/2 times the matrix D. So all I've done is I've written down what matrix B is. I'm going to do five times this matrix. So remember the rule for scalar multiplication is the scalar, the number that we're trying to multiply by multiplies every entry inside the matrix. So we get 5 * 5 is 20 five 5 * -- 1 is minus five. 5 * 1 is 5. 5 * -- 2 is minus ten. 5 * 3 is 15 and 5 * 0 is 0. So this is our answer. This is the matrix 5B or scalar five times matrix B. Notice that the matrix we get to that answer has the same size, the same order as the matrix we started from, and that's fairly obvious. That must be the case because all we do is we multiply every element inside the matrix by the scalar outside. So we are not creating any new entries in the Matrix and we're not losing any. So the matrix that we get must have the same size. So when we started with. Here's another example. We don't have to just multiply by whole numbers or two previous examples. We did 2 * A and 5 * B, but we can multiply by any number, and in this case I'm choosing to multiply the fraction or half. So I'm going to do half times matrix D. So here it is written out. Here's matrix D. We do 1/2 times that number. All we have to do is do 1/2 times every element inside the matrix. So we do 1/2 * 2, which gives us one 1/2 * 0, which gives us nought 1/2. Times 4, which gives us two. That's not the first column and 1/2 * -- 2, giving us minus 1/2 * 1, giving us a half and 1/2 * -- 1, giving us minus 1/2. And so here's our product matrix and not surprisingly, it ended up with some fractions. Then, because we were multiplying by a fraction to start off with. That concludes the video on addition and subtraction of matrices and on scalar multiplication.