 Hi and welcome to the session. Let's work out the following question. The question says using properties of determinants prove the following Determinant a, b, c, a minus b, b minus c, c minus a, b plus c, c plus a, a plus b is equal to a q plus b q plus c, q minus 3, a, b, c So let us start with a solution to this question. Let the given matrix be delta that is determinant a, b, c, a minus b, b minus c c minus a, b plus c, c plus a, a plus b Now applying c1 goes to c1 plus c2 plus c3 we get determinant a plus b plus c, b, c 0, b minus c, c minus a 2a plus 2b plus 2c c plus a, a plus b Now taking a plus b plus c common from the first column, we get a plus b plus c into determinant 1, b, c 0, b minus c, c minus a 2, c plus a, a plus b Now applying rho r3 goes to r2 minus 2r1 we get a plus b plus c into determinant 1, b, c 0, b minus c, c minus a 0, c plus a minus 2b a plus b minus 2c that is equal to a plus b plus c into b minus c multiplied by a plus b minus 2c minus c minus a into c plus a minus 2b this is equal to a plus b plus c into this gives us a b plus b square minus 2bc minus ac minus bc plus 2c square minus c square here we have c square c square plus ac minus 2bc minus ac minus a square plus 2ab or we can put a different bracket here like this this is equal to a plus b plus c into a b plus b square minus 2bc minus ac minus bc plus 2c square minus c square minus ac plus 2bc plus ac plus a square minus 2ab now we see that minus 2bc gets cancelled with plus 2bc plus ac with minus ac this is equal to a plus b plus c into now here we have a square plus b square plus c square because 2c square minus a square is c square minus a b minus bc minus ac this is equal to aq plus a b square plus ac square minus a square b minus a b c minus a square c plus a square b plus bq minus a b square minus b square c minus a b c plus a square c plus b square c plus cq minus a b c minus b c square minus ac square now we see that a b square gets cancelled with minus a b square ac square with minus ac square minus a square b with plus a square b also minus a square c with plus a square c c square b with c square minus c square b this with this and we have aq plus bq plus cq minus a b c minus a b c minus a b c is minus 3abc so this is what we are supposed to prove in this question I hope that you understood the solution and enjoyed the session have a good day