 Hello and welcome to this session. Let us understand the following problem today. Without exhaling the determinant, prove that the determinant a a square b c b b square c a c c square a b is equal to 1 a square a q 1 b square b q 1 c square c q. Now let us write the solution. Let us consider the elitist. We have the determinant a a square b c b b square c a c c square a b. Now from here, applying row operations, so applying r 1 tends to a r 1, r 2 tends to b r 2 and r 3 tends to c r 3. So we get 1 by a b c common and determinant a square a q a b c b square b q a b c c square c q and a b c. Now now taking a b c common we get a square a q 1 b square b q 1 c square c q 1 because we took a b c common from this column. So now this a b c gets cancelled with this a b c so it becomes 1 and we are left with this a square a q 1 b square b q 1 c square c q 1. Now now applying c 1 interchanges with c 3 so we get minus sign outside and determinant as 1 1 1 a q b q c q a square b square c square. Now again applying c 2 interchanges with c 3 so now we again interchanges the column c 2 and c 3 so we get again 1 more minus sign outside and we are left with determinant 1 1 1 a square b square c square a q b q c q. Now which is equal to 1 a square a q 1 b square b q 1 c square c q which is equal to RHS therefore LHS is equal to RHS hence proved. I hope you understood the problem bye and have a nice day.