CSIR NET MATHEMATICS | Linear Algebra | In-depth Solution of System of Linear Equations Dec 2017

Hello friends, in this video we are going to discuss the problem of linear algebra from CSIR NET December 2017. Example number 31 from booklet A which is of marks 3. Let A equal to 1 minus 1 1, 1 1 1 and 2 3 alpha and vector B equal to 1 3 beta. Then the system AX is equal to B over the real numbers has first option no solution whenever beta not equal to 7. Second option. An infinite number of solutions whenever alpha not equal to 2, third option an infinite number of solution if alpha equal to 2 and beta not equal to 7 and the fourth option unique solution if alpha not equal to 2. This problem is associated with system of linear equation and the options is about the solution of system of linear equation. So first we recall some important result associated System of linear equation and its solution. If AX is equal to B is a system of linear equation then it is consistent with unique solution if rank of A equal to rank of augmented matrix AB equal to N where N is number of unknowns or number of columns in A. It is consistent with infinite number of solution if rank of A equal to rank of augmented matrix AB equal to R, where R is less than N. It is inconsistent, that means system has no solution if rank of A not equal to rank of augmented matrix AB. That means this solutions, unique infinitely many solution and no solution depend on rank of A and rank of augmented matrix AB. To find rank of A or rank of augmented matrix AB, we require to convert A or augmented matrix AB into its row echelon form and number of non-zero in the row echelon form will give the rank of matrix A. So, for the given problem, we begin the augmented matrix A with the vector B. Then, we have 1, minus 1, 1, 1, 1, 1, 3, 2, 3, alpha, beta. We try to convert it into the row echelon form. So, for that, first we apply row operation such that in the first column, this two entry will become 0. So, for that, we multiply first row by minus 1, add it to the corresponding entries of second row. Multiply first row by minus 2 and add it to the corresponding entries of third row. Now, in the second step, we try to convert this 2 into the leading one. So, we multiply second row by 1 by 2. Then, we have second row 0, 1, 0, 1. Now, in the second column, all the entry below the leading one should be 0. Therefore, we multiply second row by minus 5 and add it to the corresponding entries of third row. And in the next step, we are able to get the row echelon form of augmented matrix AB. After getting the row echelon form, we are able to find the rank of AB. But rank of AB is depend on the value of alpha and beta. Note that your number of column in A is 3. Therefore, n is equal to 3. Now, the first case, if alpha equal to 2, then This last row will be 0 in A. Therefore rank of A is equal to 2 because we have two non-zero rows in the row echelon form of matrix A. Now we have two possibilities for beta. First is beta equal to 7 and second is beta not equal to 7. So if beta equal to 7 then This value becomes 0. So 0, 0 and alpha equal to 2, so 0 and 0. So in augmented matrix AB, the last row will be 0. So we have two non-zero row in the row echelon form of augmented matrix AB. So rank of augmented matrix AB equal to 2. So rank of A equal to 2, rank of AB equal to 2 and N is equal to 3. Then it is a second condition where rank of A equal to rank of AB equal to R and R less than N. Therefore system has infinite number of solution. Second case, if beta not equal to 7, that means this entry will be non-zero, hence the rank of augmented matrix AB is 3 because we have 3 non-zero row in the row echelon form of augmented matrix AB. But rank of A is 2, so rank of A not equal to rank of AB, therefore system has no solution. And the next case where alpha not equal to 2, we have 3 non-zero in the row echelon form of matrix A, therefore rank of A equal to 3. Now, 0, 0 and the non-zero entry, beta is equal to 7 or beta not equal to 7. The row echelon form of augmented matrix AB contains 3 non-zero, therefore rank of AB is 3 for any value of beta. Hence, rank of A equal to 3, rank of AB equal to 3 and 3 is equal to N. So rank of A equal to rank of AB equal to R and therefore system has unique solution. Now we are going to check the corresponding solutions. No solution whenever beta not equal to 7. So we have no solution when beta not equal to 7. It require that alpha equal to 2. So this is incomplete sentence because it does not give any value of alpha. So which is wrong. Now, the second solution, an infinite number of solution whenever alpha not equal to 2. If alpha not equal to 2, then system has unique solution. So, second option is also wrong. Now, the third option, an infinite number of solution if alpha equal to 2 and beta not equal to 7. So, alpha equal to 2 and beta not equal to 7, then system has no solution. So, third option is also wrong. And in the fourth option. Alpha not equal to 2 then system has unique solution and which must be true. Therefore this example has fourth correct answer. Thanks for watching. For more videos like share, subscribe, raise the bell icon of next gen.