 Hello everyone, Myself AS Falmari. In the last two videos we discussed the normal form and the some examples based on the normal form. Now in this video also we consider few more examples. Learning outcome at the end of the session students will be able to find the rank of the matrix using normal form. So let us consider first question reduce the following matrices to normal form and find their ranks. First problem this is one. So solution here the first entry is 3. So first we convert it to 1. So let us search for the entry 1 in the column number 1. Yes there is an entry 1 in the column number 1 okay which is in the row number 4 here. So we interchange this row number 4 and row number 1 so that this first entry will be 1. So when we interchange row number 1 and 4 at that time we will get this new matrix. This is what the matrix just we have obtained. Now our next aim is to reduce these three elements to 0 with the help of row number 1. So first to convert this 2 to 0 we will use the row transformation R2 minus 2 R1. So when we multiply row number 1 by 2 at that time we get all these entries here and now doing the subtraction of row number 1 and this resultant row we get 2 minus 2 as 0, 4 minus minus plus 2 is 6, 3 minus 4 minus 1, 6 minus of minus 6 that is plus 6 we get it as 12. Next entry is minus 1. Now it is very simple to reduce it to 0. We simply add row number 3 and row number 1. So what we will get? Obviously the first entry is 0. Second entry minus 2 plus minus 1 is minus 3, 6 plus 2 is 8, 4 plus minus 3 is 1. Now the next entry is 3. So we will perform the row transformation R4 minus 3 R1. Multiplying row number 1 by 3 we will get these entries here. So now subtracting these two rows now, 3 minus 3 is 0, 4 minus minus plus 3 is 7, 1 minus 6 is minus 5 and 1 minus minus of 9 that is plus 9 we will get it as 10. Now we will reduce the elements lying to the right side of this 1 to 0 using column number 1. As zeros are present in the column number 1 these elements are not going to change. So simply write down suitable column transformation to reduce these 3 entries to 0. First of all the entry is minus 1 and here it is 1. So we will apply a very simple column transformation C2 plus C1. So there is no change in the column number 1. In the column number 2 this element is reducing to 0 and remaining entries are as it is due to the presence of zeros here. Next entry is 2. So we will use the column transformation C3 minus 2 C1. So only the first entry is reducing to 0 and remaining are as it is. The next entry is minus 3. Now performing the column transformation C4 plus 3 times C1 what we will get? This entry is reducing to 0 and the remaining entries are as it is. Now this is what the matter is just we are having now. Now we will move towards the next diagonal entry which is 6. We will search for one entry in this second column. I think there are no ones are present in the second column. But one thing I will do I will multiply row number 2 by minus and I will add it in row number 4. At that time it will become minus 6 and 7 so the addition will become 1. So that's why I can write here by minus r2 plus r4 what I will get? When I multiply by minus to this row number 2 we will get these entries. Now adding them in row number 4 what we will get? So obviously this 0 plus 0 is 0 and minus 6 plus 7 is 1. 1 plus minus 5 is minus 4 and obviously remaining two rows are as it is. Now next we will reduce these two entries to 0. First of all minus 3. To reduce it to 0 we will use the row transformation r3 plus 3 r2. So there is no change in the first two rows. Now when we multiply row number 2 by 3 we will get these entries here now. Now adding these two so obviously first two entries are 0 minus 12 plus it is minus 4 minus 6 plus 1 is minus 5. Now the next entry is 7. We will apply the row transformation r4 minus 7 r2. So when we multiply row number 2 by 7 the entries are somewhat like this. Now making the subtraction row number 2 and row number 4 now obviously first two entries are 0. Next minus 5 minus minus plus 28 is 23 and 10 minus minus plus 14 is 24. Now next we will reduce these two elements to 0 using this column number 2. As zeros are present here so except these remaining elements are not going to change in these two columns. Now writing a suitable column transformation for this minus 4 that will be c3 plus 4 times c2 what we will get. Obviously there is no change in first two columns. In the third column the second entry is reducing to 0 and remaining three entries are as it is. Next entry is minus 2. For this we can use the transformation c4 plus 2 c2. For this again this entry is reducing to 0 here and these three entries are as it is. Now we will move towards the next entry which is minus 4. In this column and in this row one is not present but here I can observe that this is what a minus 5. So when I do the subtraction of these two columns at that time I will get here 1. For that purpose I can write here c3 minus c4. So there is no change in first two columns. Now the third column is going to change. First two entries are as it is due to the presence of zeros here. Next entry minus 4 minus minus plus 5 is 1 and 23 minus 24 is minus 1 and obviously the last column is as it is. Now next we will reduce this minus 1 to 0 using this row number 3 and which is very simple we will apply the row transformation r4 plus r3. There is no change in first three rows and in the last row first three entries are 0 obviously and this one will be what 24 minus 5 is 19. Now next we will reduce this minus 5 to 0 using this column number 3 by the transformation c4 plus 5 c3. So there is no change in the first three columns and in this last column this entry is going to reduce to 0 and remaining entries are as it is. Now finally let us move towards the next last diagonal element which is 19. So I think we can use transformation of division of a constant to the row. We can divide row number 4 by 19 so that we will get this last entry as 1 now. So there are no more diagonals. So this is what our required normal form of a given matrix. So what will be the rank? The rank of the matrix in this case is the number of nonzero rows in the normal form. Here we can count these are what the four nonzero rows. Therefore I can say the rank of the matrix is 4. Let us consider one more example solution. Let us consider one more example solution. Now here we want to reduce this 4 to 1. Now we will check for 1 in the column number 1 and row number 1. There are no ones but here this minus 3 presence it is useful for us because when I am going to add column number 1 in column number 2 at that time this first entry will be 1. For that purpose I can write by performing the transformation C1 plus C2 what we get. There is no change in last two columns. Now the column number 1 will be what? 4 minus 3 is 1, 12 minus 9 is 3 and 20 minus 15 is 5. Now next we will reduce these two elements to 0. For that purpose we will apply two row transformations okay. What are that two row transformations? First of all by R2 minus 3 R1. Okay what we get now here? So when I multiply row number 1 by 3 so we will get these three entries now making the subtraction 3 minus 3 will be 0 minus 9 minus minus plus 9 will be 0 and 18 minus 18 is 0. Now pause this video and write down a row transformation to reduce this 5 to 0. I hope that all of you have written the answer. The row transformation to reduce this 5 to 0 will be what? R3 minus 5 R1. So when we multiply row number 1 by 5 we get these entries now here. Now subtracting 5 minus 5 is 0 minus 15 minus minus plus 15 is 0 and 30 minus 30 is 0. Now we will reduce these two elements to 0 now using column number 1 only. First of all minus 3 for that we will apply the column transformation C2 plus 3 C1. So what we get? There is no change in the column number 1. When we multiply column number 1 by 3 we will get these entries and when we do the addition of these two columns this entry is reducing to 0 and remaining entries are as it is. Next entry is 6 so we perform the transformation C3 minus 6 C1. So when I multiply column number 1 by 6 we get the entries like 6 0 0 and taking the subtraction it will reduce this to 0 and remaining entries are as it is. Now except this first entry all are 0s it means this is what our required normal form. So in this case the rank of the matrix will be the number of non-zero rows. I think there is only one this non-zero row is present that's why the rank of the matrix is 1.