#csirnet #mathematics #linearalgebra FEB 2022 | Singular Matrices

Hello friends, this is Mahesh Yoylekar and you are watching Next Gen Maths. If you have not subscribed the channel, please subscribe the channel and press the bell icon. In this video, we are going to discuss the problems assigned at FEB 2022 Problem of Linear Algebra. The question is, let A and B be n x n matrices. Suppose the sum of element of any row of A is 2, sum of any column of B is 2, which of the following matrices necessary singular. So we are given four options, first option is I-1 by 2 B into A transpose, second option is I-1 by 2 into AB, third option is I-1 by 4 into AB, fourth option I-1 by 4 into B into A transpose. Here we want to change the singularity of this given matrices. To solve this example, we will use counter example. Consider two matrices A equal to 1, 1, 2, 0 and B equal to 1, 2, 1, 0. The condition for A is given in the data. Sum of element of any row of A is 2, so matrix A's first row is 1, 1 and its sum is 2 and second row is 2, 0, its sum is also 2, so it satisfy this condition. And for matrix B's condition is sum of any column of B is 2, so first column of matrix B is 1, 1 and its sum is 2 and the second column is 2, 0, its sum is also 2, so B is also satisfy the condition. Now we multiply A with B, so AB equal to 1, 1, 2, 0 multiplied 1, 2, 1, 0 which will give 2, 2, 2, 4. Next we take the multiplication of B with A transpose, so 1, 2, 1, 0 multiplied transpose of A 1, 2, 1, 0 and we will get 3, 2, 1, 2 and 2 cross 2 identity matrix is 1, 0, 0, 1. Now we will check all the option one by one for its singularity. So option one, I minus 1 by 2 into B into A transpose, so 1, 0, 0, 1 minus 1 by 2 into B into A transpose is 3, 2, 1, 2, multiply 1 by 2 with the matrix then we have 3 by 2, 1, 1 by 2, 1 minus 3 by 2 is minus 1 by 2, 0 minus 1 is minus 1, 0 minus 1 by 2 is minus 1 by 2 and 1 minus 1 by 2 is 0. We are interested in the singularity, so we want to find the determinant value, so determinant I minus 1 by 2 B into A transpose equal to minus 1 by 2 which is non-zero. Determinant is non-zero, so it is a non-singular matrix, so option one is not in correct answer. Option number two which is I minus 1 by 2 AB, so 1, 0, 0, 1 minus 1 by 2 into AB is 2, 2, 2, 4, so 1 by 2 multiply with this entries, we have 1, 1, 1, 2, so 1 minus 1 is 0, 0 minus 1 is minus 1, 0 minus 1 is minus 1 and 1 minus 2 is minus 1. Now as determinant value is minus 1 which is non-zero, therefore I minus 1 by 2 into AB is a non-singular matrix, so option number two is not the correct option. Now option number three which is I minus 1 by 4 into AB, so 1, 0, 0, 1 minus 1 by 4 into AB is 2, 2, 2, 4, now multiply 1 by 4 with 2, 2, 2, 4, we have 1 by 2, 1 by 2, 1 by 2, 1, 1 minus 1 by 2 is 1 by 2, 0 minus 1 by 2 is minus 1 by 2, 0 minus 1 by 2 is minus 1 by 2 and 1 minus 1 is 0. If you find the determinant value, then we have minus 1 by 2 which is non-zero, so option number three which is I minus 1 by 4 AB is also non-singular, so the option number three is not a correct option. Fourth option which is I minus 1 by 4 into B into A transpose, so 1, 0, 0, 1 minus 1 by 4 into 3, 2, 1, 2, 1 by 4 multiply with this entry, we have 3 by 4, 1 by 2, 1 by 4, 1 by 2, so 1 minus 3 by 4 is 1 by 4, 0 minus 1 by 2 is minus 1 by 2, 0 minus 1 by 4 is minus 1 by 4 and 1 minus 1 by 2 is 1 by 2, if you find the determinant value, then which is 1 by 8 minus 1 by 8 equal to 0, determinant of I minus 1 by 4 into B into A transpose is 0, so it is a singular matrix, therefore option number four is the correct option. Thanks for watching. For more videos like, share, subscribe, press the bell icon of NextGenMS.