CSIR NET MATHEMATICS | Linear Algebra | Determinant of Matrix DEC 2011

Welcome friends, in this video we are going to discuss the problem of Linear Algebra from C.S. Ionide December 2011. Example, the determinant of matrix is, we have four options 0, minus 9, minus 27, plus 1. Evaluate the determinant we have two methods, first method is cofactor method and second one is row operation. For the higher order matrices row operation method is more effective than the cofactor method and the given matrix is of order 6 cross 6. Therefore, row operation method is more effective than the cofactor method. Now, before applying row operation, first we recall the effect of row operation on the determinant value. Basically, we have three row operations. First row operation is row interchange operation. After the row interchange operation, the determinant of A star, A star is a matrix obtained after applying the row operation on a has negative value of the determinant of a second operation if we multiply any row by some scalar then the determinant of a star is k times the determinant of a and the third operation if we multiply one row by some scalar and add it to the another row then the determinant value will remain same so determinant of a star equal to a so we will use this result to evaluate the determinant now for the given matrix we will apply three operation first row operation r16 minus 2 r25 minus 2 r34 minus r16 minus 2 means first row multiply minus 2 added to the sixth row then minus 2 plus 2 cancel so we have 0 and minus 4 plus 1 we have minus 3. Now for the second row operation second row is multiply minus 2 added to the fifth row so minus 2 plus 2 0 and minus 4 plus 1 minus 3. Similarly third operation third row multiply by minus 2 added to the fourth row so minus 2 plus 2 0 and minus 4 plus 1 minus 3. Now this matrix A star is obtained by applying the row operation of type 3 where one row is multiplied by some scalar added to the third row and therefore the determinant value will remain same. So determinant of a equal to determinant of a star and it is a triangular matrix and for the triangular matrix the determinant value is nothing but the multiplication of diagonal values which are 1 1 1 minus 3 minus 3 minus 3 so 1 into 1 into 1 minus 3 into minus 3 into minus 3 which is minus 27 and which is option number 3 so option number 3 is correct answer thanks for watching for more videos like share subscribe please the bell icon of next gen math