CSIR NET MATHEMATICS| Linear Algebra | Theory, Results and Solution of Linear Problem JUNE 2018

In this video, we are going to discuss the problem of linear algebra from CSIRnet June 2018. Example number 72. Booklet A which is of mark 4.75. The problem is if A is a 5 cross 5 matrix and the dimension of the solution space of Ax is equal to 0 is at least 2 then option 1 rank of a square less than equal to 3, second option rank of a square greater than equal to 3, Third option, rank of x square equal to 3 and determinant of a square is equal to 0. To solve this example, we begin with some definitions and some results. So, first definition, rank of a. So, dimension of row space of a is called rank of matrix a. Second definition, nullity of matrix a. The dimension of solution space ax is equal to 0. A is a nullity of matrix A. Then on base of these two definition we have one theorem that is called the rank nullity theorem or dimension theorem. If A is a matrix of order M cross N then rank of A plus nullity of A equal to N. N means total number of columns in A. Now for the given problem we have given that A is a 5 cross 5 matrix. So N is equal to 5 which is the number of columns. And we have given at the dimension of the solution space of AX which is called the nullity and which is at least 2. So, nullity of A is greater than equal to 2. Now, if we use this theorem rank of A plus nullity of A equal to n. So, nullity of A is equal to n minus rank of A. So, n minus rank of A which is also greater than equal to 2. Therefore, n equal to 5. So, 5 minus rank of A greater than equal to 2. Hence, we have 3 greater than equal to rank of A or you can rewrite in the form of rank of A is less than equal to 3. Now, we have some results. So, first result rank of product of A B is less than equal to rank of A. So, if B is equal to A that means rank of A into A is also less than equal to rank of A. Therefore, rank of A square less than equal to rank of A. We already proved rank of A is less than equal to 3. So, rank of A square is also less than equal to 3. So, the first option is rank of A square is less than equal to 3, which is correct. Now, rank of A square greater than equal to 3, which is wrong. And in the third option, we have exact equality, which is also not possible. Now, for the fourth option, we will recall some results of determinant. So, first result, let A be a square matrix of order n. Then A is non-singular that means determinant of A not equal to 0 if and only if rank of A equal to N. Since N is equal to 5 and rank of A is less than equal to 3 therefore rank of A must be 0. Because result says that A is non-singular that means determinant A not equal to 0 if and only if rank of A is equal to N. N is equal to 5. But rank of A is less than equal to 3, which is less than 5. Therefore, determinant of A equal to 0. Now, we have a property of determinant. Determinant of A square is equal to determinant of A into determinant of A. So, 0 into 0 is equal to 0. Therefore, determinant of A square equal to 0. And hence, the fourth option is correct. So, this example has only two options correct, first and fourth. Thanks for watching. Like. Share, subscribe, raise the bell icon of NextGenMess.