 Hello everyone, myself Ayaz Falmari. In the last video we discussed the normal form to find out the rank of a matrix and also considered an example. Now in this video also we consider few more examples. Learning outcome, at the end of this session students will be able to find the rank of the matrix using normal form. First, reduce the following matrix to normal form and find its rank. Okay, this is what the first matrix solution. Now we know that in the normal form this first entry is 1. So we will search for the entry 1 in the first column. Yes, there are 2 entries which are present in the row number 3 and row number 5. Now I will simply interchange row number 1 and row number 3. So obviously the remaining rows are as it is. Now interchanging row number 1 and 3 then we will get this new resultant system. Now once we convert this first entry to 1, now the next aim is to convert the elements lying below to this row number 1 to 0 using this row number 1 only. So as minus 3 is present, so we will perform the row transformation R2 plus 3 R1. Now here only the row number 2 is going to change for the first transformation copying the row number 1 as it is. So when we multiply by 3 to the row number 1 we will get these entries. So obviously the first entry is 0. Next entry 3 minus 6 is minus 3, minus 3 plus 12 is 9, minus 3 plus 9 is 6, minus 3 plus 15 is 12. Now next entry is 2. We perform the transformation R3 minus 2 R1. Multiplying row number 1 by 2 then we will get these entries. Now performing the transformation 2 minus 2 is 0, 6 minus of minus 4 that is plus 10, minus 2 minus 8 minus 10, 6 minus 6 0 and 10 minus 10 0. Now next again 2 is present. So we will perform the transformation R4 minus 2 R1. So when I multiply row number 1 by 2 then we will get all these entries. So obviously the first entry is 0, next entry is 0, minus of minus that is plus 4, 4 minus 8 that is minus 4, 6 minus 6, 0 and 10 minus 10, 0. Now the last entry is 1. So we will perform the simple row transformation R5 minus R1. So first entry obviously is 0, second entry is 0, minus of minus 2 that is plus 2, 2 minus 4 minus 2, 3 minus 3, 0 and 5 minus 5, 0. Now next we will convert the element line to the right side of this 1 to 0 with the help of this column number 1. So first of all to bring 0 here, so we will use the column transformation C2 plus 2 C1. So obviously there is no change in the column number 1, so I can copy this column number 1 and obviously there is no change in this block also. Now when I multiply column number 1 by 2 we will get these entries. Now adding these two columns only first entry is going to change and remaining entries are as it is due to the presence of zeros here and it is going to happen for this total block. Next entry is 4, so we perform the transformation C3 minus 4 C1 multiplying column number 1 by 4 we will get all these entries and performing the transformation, first entry reducing to 0 remaining entries are as it is. Now next entry is 3, so we perform the transformation C4 minus 3 C1 multiplying column number 1 by 3 and performing this transformation here now. This first entry is 0 now and the remaining entries are as it is. And finally this last entry is 5, so we perform the transformation C5 minus 5 C1. When I multiply column number 1 by 5 I will get all these entries and when I perform this transformation this first entry reducing to 0 and the remaining entries are as it is. Now we will move towards the next diagonal which is minus 3. We search for one in the second column and in the second row obviously there is no entry. Now we select a suitable row or a column. I think this entry is 4 is good for this entry minus 3 because their addition is 1. So what I will do now here? I will use the transformation R2 plus R4, okay only row number 2 is going to change and remaining rows are as it is. Adding the row number 2 and row number 4 first entry is 0, second entry minus 3 plus 4 is 1, next entry 9 minus 4 is 5, 6 plus 0 is 6, 12 plus 0 is 12. Now next we convert the elements lying below to this one entry is 0 with the help of row number 2 only. So first due to the presence of 10 we will use the row transformation R3 minus 10 R2, okay first two rows are as it is. Now the third, fourth and fifth row is going to change. First of all third row is changing. Multiplying row number 2 by 10 then we will get all these entries. So obviously first two entries are 0, next entry minus 10 minus 50 is minus 60, 0 minus 60 minus 60 and 0 minus 120 is minus 120. Now next entry is 4. So we will perform the transformation R4 minus 4 R2. Multiplying row number 2 by 4 I will get these entries. Obviously first two entries are 0, next entry minus 4 minus 20 minus 24, 0 minus 24 is minus 24 and 0 minus 48 is minus 48. And this last entry is 2. So we perform the transformation R5 minus 2 R2. Multiplying row number 2 by 2 and performing this transformation obviously first two entries are 0. Now the third entry minus 2 minus 10 is minus 12, 0 minus 12 is minus 12 and 0 minus 24 is minus 24. Now we will convert the elements lying to the right side of this 1 to 0 using this column number 2 only. So first due to the presence of 5 we will use the transformation C3 minus 5 C2. Due to the presence of 0s in these places these remaining entries are not going to change only this 5 is reducing to 0 ok. That's why I can write 0, 0 and these remaining entries are as it is. Next entry is 6. So we will use the transformation C4 minus 6 C2. Due to this transformation this entry is converting to 0 and the remaining entries are as it is. And finally this entry is 12. We use the transformation C5 minus 12 C2 and obviously this 12 is reducing to 0 and the remaining entries are as it is due to the presence of 0s here. Now I will divide the third row by minus 60, fourth row by minus 24 and fifth row by minus 12 due to the presence of the elements in multiples of minus 60, minus 24 and minus 12. So first when I divide the third row by minus 60 then I will get these entries ok. After dividing fourth row by minus 24 we will get these entries now here and finally when we divide last row by minus 12 we get these new entries ok. So now we will move toward the next entry. So obviously the next entry is 1. Now we will convert the elements lying below to this 1 entry to 0 using this row number 3 only. So due to the presence of 1 and 1 here we will use the simple row transformations R4 minus R3. So obviously first three rows are as it is due to the identical rows this row is totally vanishes. And for fifth row we will use the transformation R5 minus R3. Again row number 3 and row number 5 are identical therefore the last row is also vanishes. Now we will convert the elements lying to the right side of this 1 to 0 using this third column only. So due to the presence of 0s here remaining entries are not going to change simply write down a suitable column transformation C4 minus C3. So there is no change in the first three columns. In the fourth column this entry is going to reduce to 0 and remaining are as it is. And for last column we will consider a transformation C5 minus 2 C3. Again this 2 is reducing to 0 and the remaining entries are as it is. So when I move toward the next diagonals the next diagonals are 0s and the elements in that row and column are also 0s. So this is what our required normal form of a given matrix. So we will count simply the nonzero rows present in this normal form. So here are 1, 2, 3 and these remaining two are the zero rows we will not take them. So the rank of the matrix is what? 3.