 Hello and welcome to the session. In this session we discuss the following question which says, what is the value of the determinant with elements in the first row as 0 to 0, in the second row as 2, 3, 4 and in the third row as 4, 5, 6. Let's proceed with the solution now. We suppose let delta be equal to the determinant with elements 0 to 0 in the first row, in the second row as 2, 3, 4 and in the third row as 4, 5, 6 and we are supposed to find the value of this determinant. So this would be equal to this first element of the first row that is 0 into the determinant with elements 3, 4 in the first row. This is the first element of the first row and the first column. So we would not take any other elements from the first row and the first column and instead we would take these remaining elements into this to be multiplied by this first element. Now then we take the second element of the first row which is 2 and we put a minus sign before 2 and we have minus 2 and this is to be multiplied in the second row. Now consider this row containing this element and this column containing this element. Now consider this third element of the first row which is 0. So here we have plus 0 into the determinant with elements 2, 3 in the first row. Now the 0 multiplied by this determinant would give us 0 itself and this would be equal to multiplied by 6 is 12 minus 4 multiplied by 4 which is 16 plus 0 multiplied by this determinant is 0. And so this is equal to minus 2 into which is equal to tau is equal to 8 and we had assumed this delta to be the given determinant. The value of the given determinant with elements 0, 2, 0 in the first row, 2, 3, 4 in the second row, 4, 5, 6 in the third row is equal to this complete C session. Hope you have understood the solution of this question.