 Hello and welcome to the session. In this session we discuss the following question which says by using the properties of determinants show that determinant with elements a plus b, b plus c, c plus a, b plus c, c plus a, a plus b, c plus a, a plus b, b plus c is equal to 2 times the determinant with elements a, b, c, b, c, a, c, a, b. Let's move on to the solution now. We need to show that determinant with elements a plus b, b plus c, c plus a, b plus c, c plus a, a plus b, c plus a, a plus b, b plus c is equal to 2 times determinant with elements a, b, c, b, c, a, c, a, b. We consider the LHS that is determinant with elements a plus b, b plus c, c plus a, b plus c, c plus a, a plus b, c plus a, a plus b, b plus c. We take let this be equal to delta. Now we apply c1 goes to c1 plus c2 plus c3 and then applying this we have delta is equal to determinant with elements 2 into a plus b plus c, b plus c, c plus a, 2 into a plus b plus c, c plus a, a plus b, 2 times determinant. into a plus b plus c a plus b b plus c that is the elements of the column c2 and c3 remain the same but the elements of column c1 are given by c1 plus c2 plus c3. Let's recall one property which says that it's given delta equal to a determinant with elements a1 b1 c1 a2 b2 c2 a3 b3 c3 and we multiply a column of this determinant by a constant say k that is you multiply each element of this column 1 by a constant k then we can take out this common factor from the column so this would be equal to k into the determinant with elements a1 b1 c1 a2 b2 c2 a3 b3 c3 so we'll use this property here and so we'll take out this constant 2 from the first column outside so we would get delta is equal to 2 times determinant with elements a plus b plus c b plus c c plus a a plus b a plus b plus c a plus b b plus c next we apply c2 goes to c1 minus c2 and on applying this we have delta is equal to 2 times determinant with elements a plus b plus c a c plus a a plus b plus c b a plus b a plus b plus c c b plus c that is the elements of the column c1 and c3 remain same but the elements of c2 are given by c1 minus c2 next we apply c3 goes to c1 minus c3 and on applying this we have delta is equal to 2 times determinant with elements a plus b plus c a b a plus b plus c b c a plus b plus c c a that is the elements of columns c1 and c2 remain same and elements of c3 are given by c1 minus c3 then next we apply c1 goes to c1 minus c2 minus c3 and on doing this we have delta is equal to 2 times determinant with elements c a b a b c b c a next we recall one more property which says that if any 2 columns or rows of a determinant are interchanged then sign of the determinant changes. So here we interchanged the columns c2 and c3 so we get delta is equal to 2 into minus 1 into determinant with elements c b a a c b b a c next again we interchanged the columns c1 and c3 so we would get delta is equal to 2 into minus 1 square that is again we would multiply it by minus 1 since we know that the sign of the determinant changes so this into determinant with elements a b c b c a c a b so this means we now get delta is equal to 2 times determinant with elements a b c b c a c a b and this is equal to the RHS so thus we get the LHS is equal to the RHS hence proved. So this completes the session hope you have understood the solution for this question.