 In this video, we are going to learn a method to find the rank of a matrix. At the end of this session, students will be able to find the rank of the matrix using normal form. Let us see what is normal form or canonical form. Every m cross n matrix of rank R can be reduced to the form like this by a finite sequence of elementary transformations. This form is called the normal form or the first canonical form of the given matrix A. Okay, whatever the given matrix of order m cross n is there, that can always be reduced to this form. Okay, this form has a special characteristic. Here we can see that now here is one block of elements and that are looking like somewhat equal to the identity block and remaining are the zero blocks. This matrix in short can be denoted by this symbol IR 000. This IR is what this is what the identity block of order R. And these three zeros are the zero blocks of various orders. So, each and every matrix of rank R can always be reduced to such a form and such a form is called what the normal form or the canonical form. Working rule to reduce the given matrix to normal form. Step number one, suppose we have given this matrix. In this normal form, we observe that the first entry will be one. Now here it is A11. So, in the step number one, our aim is to convert this A11 to one if it is not there. Okay, how to convert it? We can choose two suitable rows or columns so that this entry will be one. And if this is not possible, then we are going to subtract a suitable column or a suitable row from the first row or a first column such that this A11 is reduced to one. And one more way to convert this A11 to one is by dividing the first row or a first column by A11. So, the first aim is to convert this element A11 to one. Okay, if I convert it by using the row transformations, then the resultant matrix will look like this one A12 dash A13 dash and so on. These dashes are indicating the elements are changed once. Step number two, this is what the matrix we have obtained from step number one. Now, our next job is to convert the elements lying below to this one to zero with the help of this row number one. And after that, our aim is to convert the elements lying to the right of this one to zero with the help of the column number one. To do this, we have to perform the row and column transformations of the type R2 plus minus K R1 and C2 plus minus J C1. Once we convert the remaining elements lying in the first row and first column to zero with the help of first row and first column, the resultant matrix will somewhat looks like this and the dashes in these elements are representing that elements are changed. Now the step number three, now we have to move towards the next diagonal, which will be A22 dash, then we'll convert this element to one. If we convert this element to one, then we'll convert the remaining elements lying below and to the right side of this entry to zero with the help of the row number two and the column number two, and we'll continue the same procedure until the last diagonal. So once I perform this step number one, two, three repeatedly for each diagonal, finally, I will get the resultant matrix is somewhat like this. And this is what I will require the normal or the canonical form. Once I get the normal or a canonical form, in this form, the rank of the matrix is defined as the number of non zero rows present. Okay, simply you have to count the number of non zero rows. That will be the rank of that corresponding matrix. Let us consider one question. Reduce the following matrix to normal form and hence find its rank solution. Now in this given matrix, this first entry is two. Now our aim is to convert it to one. First I will search for the entry one in the first column. Yes, there is one. Now simply we interchange row number one and row number three. So the middle row as it is now interchanging the first and last row, then we will get this new resultant matrix. Now the next aim is to convert the elements lying below to this one to zero using this row number one. To convert these two elements to zero, we perform the row transformations R2-3 R1 and R3-2 R1. Obviously, there is no change in the first row. Now performing first of all this transformation R2-3 R1, multiplying row number one by three, then we'll get all these entries. Now subtracting row number two and this row number one, what we will get? This is three minus three is zero minus three minus three minus six, one minus three minus two and two minus six is minus four. Now next performing this transformation R3-2 R1, multiplying row number one by two, we will get these entries. Now performing this transformation two minus two zero, one minus two minus one, minus three minus two minus five, minus six minus four minus ten. Now we'll convert the elements lying to the right side of this one to zero, using this column number one. Now as zeros are present, so obviously this block is not going to change. Okay, simply we will write a suitable column transformation to reduce these entries to zero and we'll copy this block as it is, as they are not going to change due to the presence of zeros here. Okay, this is what the matrix in the last step. Now we'll perform these column transformations C2 minus C1, C3 minus C1 and C4 minus two C1. So obviously there is no change in this much block. Now only these three elements are reducing to zero. Now I will move towards the next diagonal which is minus six, but we want here one, but below to this minus six minus one is present. So first I will interchange row number two and row number three. Then I will get this new resultant matrix here. I will multiply row number two by minus one, so that this minus one is converted to one. So first as it is, if we multiply by minus to the second row, second row will be zero one five ten and the third row as it is. Now pause this video and write down a row transformation to reduce this minus six to zero with the help of row number two and also calculate that corresponding row. I hope that all of you have written the answer. The row transformation to reduce this minus six to zero with the help of row number two will be R3 plus six times R2. So row number one as it is, obviously row number two also as it is. Now we are multiplying by six to the row number two, then at that time we will get these entries. Now performing the addition, obviously first two entries are zero. Next two entries minus two plus 30 is 28 minus four plus 60 is 56. Now verify your answers. Now next we will reduce the elements line to the right side of this one to zero with the help of this column number two. So obviously due to the presence of zero in the first and last position, only these two entries are going to change. Okay, so write a suitable column transformation so that these two entries are zero. Now the required column transformations are what? C3 minus five times C2 and C4 minus 10 times C2. So apart from these two entries, the remaining entries are as it is. Now these two entries are zero. Now here next diagonal entry is 28. So we'll divide third row by 28 here. Then we will get the first two rows as it is and the last row will be 0012. Now this is automatically converted to one. Now simply we have to convert this two to zero with the help of column number three. Now here we perform the column transformation C4 minus two times C3. So obviously remaining entries as it is these two is only converted to zero. All the diagonals are finished. So this is what our required normal form of a given matrix. So now we are capable to find out its rank. So the rank will be number of non-zero rows. How many non-zero rows are there? Let us count one, two and last one three. So the rank of the matrix is three.