 Hi, and welcome to the session. Let us discuss the following question. The question says, show that the determinant of order 3 in which elements are x plus 1, x plus 2, x plus a, x plus 2, x plus 3, x plus b, x plus 3, x plus 4 and x plus c is equal to 0, where a, b, c, r and a, p. Now, begin with the solution. Since a, b and c and a, p plus a is equal to c minus b, this implies 2b is equal to a plus c. Now, we will first consider the left-hand side. Left-hand side is equal to determinant of order 3 in which elements are x plus 1, x plus 2, b and x plus c. Column operations. The first column operation that we will apply is c2 goes to c2 minus c1. Second column operation that we will apply is c3 goes to c3 minus c1. So, let us now apply these column operations. In the first column, elements will remain as it is, that is, x plus 1, x plus 2 and x plus 3. In second column, first element will be x plus 2 minus x minus 1, that is equal to 1. Second element will be x plus 3 minus x minus 2, that is 1. Third element will be x plus 4 minus x minus 3, that is 1. In the third column, we will have x plus a minus x minus 1, that is a minus 1. Second element will be x plus b minus x minus 2, that is b minus 2. Third element will be x plus c minus x minus 3, that is c minus 3. Now, we will apply row operation to this determinant. First row operation that we will apply is r2 goes to r2 minus r1. In second row operation that we will apply is r3 goes to r3 minus r1. So, let us now apply these operations to this determinant. First row will remain as it is. In second row, we will have x plus 2 minus x minus 1, that is 1. Then we have 1 minus 1, that is 0, b minus 2 minus a plus 1, that is b minus a minus 1. In third row, we will get elements as x plus 3 minus x minus 1, that is 2, 1 minus 1, that is 0, c minus 3 minus a plus 1, that is c minus a minus 2. We will learn the theorem that says that if two rows or columns of a determinant are interchained, then the sign of the determinant is j. So, by interchanging in second column, determinant of 1, 0, 0, 1, 1, 2, a minus 1, b minus a minus 1, c minus a minus 2. So, we equal to minus into 1 into determinant of 1, b minus a minus 1, 2, c minus a minus 2, minus x plus 1 into determinant of 0, 0, b minus a minus 1, c minus a minus 2, minus 1 into determinant of 0, 0, b minus 2. 1, 2 is equal to e to the determinant of 1 e minus a minus 1 minus a minus 2 into c minus a minus 2 e minus a minus 1. Now this is equal to minus a minus 2 minus 2b plus 2 is equal to minus into minus 2b. So by substituting 2b in place of c plus a we get minus 2b 0. And this is equal to right hand side. So we have proved that left hand side is equal to sum each cessation by and again.