CSIR NET MATHEMATICS JUNE 2018 | Linear Algebra | Solution of Section B

In this video, we are going to discuss the example of Linear Algebra from CSIRnet, June 2018. Let A be a m x n matrix and B be a n x m matrix over the real numbers with m less than n. Then the options are A is always non-singular, A is always singular, B is always And fourth, BA is always singular. Case 1, suppose A is 2 cross 3 matrix AB 0 CD 0 and B is 3 cross 2 matrix 0000 EF. If we multiply AB then it is resultant matrix is 2 cross 2 and its value is 00. Therefore, AB is a singular matrix. If we take A is equal to I am. That means identity matrix of order M and 0 is M cross N minus 1. That means remaining M cross N minus 1 entries are 0. So A is a M cross N matrix and B where first M cross M matrix are the identity and the remaining values are 0 which is a N cross M matrix. If we multiply A and M. Then, AB is equal to IM which is the identity matrix of order M and which is also non-singular. Therefore, we cannot say AB is always non-singular or AB is always singular. So, first two answers are cancelled. Case 3, A is a M cross N matrix with M less than N. So, the reduced row echelon form of matrix A contains at most minimum of M and N leading number of 1s. So minimum is m, so therefore matrix A contains at most m leading ones which is less than n, n is the number of columns. Therefore the system of linear equation Ax is equal to 0 has non-zero solution because rank of A, rank of A colon 0 is less than n. So, this is the condition of system of linear equation which shows that system as So, this homogeneous system may have non-zero solution. Therefore, BAX, multiply this equation both side by B, matrix B. So, BAX equal to 0 has also non-zero solution and so BA is singular matrix. So, BA is always singular matrix. So, this is our final answer. Let ABA 2x2 matrix over R With determinant A plus I is equal to 1 plus determinant A, then we can conclude that option A, determinant A is 0, A equal to 0, trace of A is 0, 4, A is non-singular matrix. Answer 28. A is equal to A, B, C, D, 2 cross 2 matrix. A plus I is equal to 1 plus A, B, C, 3D plus 1. Now we are given determinant A plus I is equal to 1 plus determinant A. So determinant A plus I is equal to A plus I B C D plus 1 equal to 1 plus A B C D. So determinant of A plus I is equal to 1 plus A into D plus 1 minus B C equal to 1 plus A D minus B C. Now, by opening the brackets, a plus 1, d plus 1. We have AD plus A plus D plus 1 minus BC. Therefore, these terms are cancelled. AD cancelled with the AD. BC cancelled with BC. And 1 cancelled with 1. So, we have remaining part. A plus D is equal to C. Which is a trace of A. So, trace of A is equal to 0. Therefore, final answer is trace of A equal to 0. Example number 30. The trace of matrix 210. 0 2 0 0 0 3 power 20 is option first 7 power 20 option 2 2 power 20 plus 3 power 20 option 3 2 into 2 power 20 plus 3 power 20 option 4 2 power 20 plus 3 power 20 plus 1 answer a equal to 2 1 0 0 2 0 0 0 3 so we will use the result stress of matrix a So, we will try to find the eigenvalues of matrix A. The eigenvalues of matrix, upper triangular matrix A are the diagonal values which is 2, 2, 3. Now, another result, the eigenvalues of A power 20 are the power of 20 power of eigenvalues of A. So, 2 power 20, 2 power 20 and 3 power 20. So, therefore, the trace of A power 20, which is the required matrix, 2 power 20 plus 2 power 20 plus 2 3 power 20 the sum of their eigenvalues is equal to 2 into 2 power 20 plus 3 power 20. So final answer is 3. Example let Rn n greater than equal to 2 be equipped with standard inner product let v1 v2 vn be an column vectors forming an orthonormal basis of Rn. And let A be n cross n matrix formed by the column vectors V1, V2, Vn. Then the options are A equal to A inverse, second A equal to A transpose, third A inverse is equal to A transpose and the fourth determinant A equal to 1. Now, we have given V1, V2, Vn form a orthonormal basis for Rn. Therefore, we have property Vi transpose dot Vj. That means dot product or inner product between vi and vj is equal to 1 if i equal to j. Otherwise, it is 0. That means i not equal to j, value is 0. Now, we are constructing matrix A with the column vectors v1, v2, vn. Therefore, A transpose dot A is equal to v1, v2, vn transpose dot v1, v2, vn. So, after the multiplication we have n cross n matrix in which the first element is v1 transpose dot v1, second element is v1 transpose dot v2 and the nth element of the first row v1 transpose dot vn. Similarly second row is v2 transpose dot v1, second element of the second row v2 transpose dot v2 and the nth element of second row v2 transpose dot vn and similarly the all the other rows. The given V1, V2, Vn are the orthonormal basis. Therefore, V1 transpose dot V1 is equal to 1 and other entry on the first row are 0. Now similarly, in the second row, V2 transpose dot V2 will give the value 1 and the other value will be 0. And similarly, in the last row, Vn transpose dot Vn is 1 and the other entries are 0. So, finally, it convert into the identity matrix of order n cross n. Therefore, A transpose dot A is equal to identity. So, you can write in the form A inverse is equal to A transpose. Hence, the third option A inverse is equal to A transpose is correct. Now, to check the other options, we are going to use the counter example. The counter example A is equal to 1 by root 2 minus 1 by root 2, 1 by root 2 plus 1 by root 2. Now, v1 equal to first column 1 upon root 2, 1 upon root 2 and v2 is equal to minus 1 by root 2 plus 1 by root 2. So, norm of v1 and v2 is 1 because 1 by 2 plus 1 by 2 equal to 1 and v1 dot v2 or inner product between v1 and v2 is 0. So, minus 1 by 2 plus 1 by 2 will be 0. So, it satisfy the given condition of the example. a transpose is equal to a inverse is equal to 1 by root 2 minus 1 by root 2, 1 by root 2, 1 by root 2 by singly taving transpose of a but which is not equal to a. Therefore, the first option a is equal to a inverse which is wrong and the second option a equal to a transpose which is also wrong. Now, for the fourth option, we are taking another counter example. a is equal to 1 0, 0 minus 1. It's also satisfy the given condition. First column vector 1 0 as norm 1 and second column vector 0 minus 1 as also norm 1. And the dot product between 1 0 and 0 minus 1 is 0. Therefore v1 and v2 are the orthonormal basis. Now the determinant of a is equal to minus 1. Therefore the fourth option we say the determinant equal to 1 which is wrong. Only third option is correct. A inverse is equal to A transpose. Thanks for watching. For more videos, like, subscribe, press the bell icon of Next Gen Maths.