 Hello and welcome to this session. Let us understand the following problem today. Using cofactors of element of second row evaluate delta is equal to 53a201123. Now let us write the solution. Now first let us find the cofactor of the second row. So cofactor of a21 which is this element is equal to minus 1 to the power 2 plus 1 m21 which is equal to minus 1 cube into determinant 3a23 which is equal to minus 1 into 9 minus 16 which is equal to 7 and which is our a21. Now cofactor of a22 is equal to minus 1 to the power 2 plus 2 m22 which is equal to minus 1 to the power 4 into determinant 5813 which is equal to 15 minus 8 which is equal to 7 and its our a22. Now cofactor of a23 is equal to minus 1 to the power 2 plus 3 m23 which is equal to minus 1 to the power 5 into determinant 5312 which is equal to minus 1 into 10 minus 3 which is equal to minus 7 which is our a23 therefore now we will evaluate the determinant therefore it is equal to a21 a21 plus a22 a22 plus a23 a23 where small a is the element and capital A is the cofactor now which is equal to 2 into 7 plus 0 into 7 plus 1 into minus 7 which is equal to 14 minus 7 which is equal to 7 therefore required answer is 7. I hope you understood the problem. Bye and have a nice day.