 Hello and welcome to the session. In this session we discussed the following question with safe by using properties of determinants, prove the following. Determinant with elements x plus 4, 2x, 2x in the first row. When in the second row the elements are 2x, x plus 4, 2x. In the third row the elements are 2x, 2x and x plus 4 is equal to 5x plus 4, the whole multiplied by 4 minus x the whole square. Before we move on to the solution let's recall one property of determinant according to which we have that if each element of a row or a column of a determinant is multiplied by a constant k then its value gets multiplied by k. And by using this property of determinant we can take out any common factor from any one row or any one column of the given determinant. So this is the key idea that we would use in solving this question. Let's move on to the solution now. First of all we take when delta be equal to the given determinant. So this is the given determinant delta. Now we apply one operation. So applying the operation C1 goes to C1 plus C2 plus C3 we get delta is equal to now we will represent the elements of the column 1 C1 that is elements of this column are represented by C1 plus C2 plus C3. So now the first element of the column 1 that is C1, x plus 4 would be written as x plus 4 plus 2x plus 2x. So that would become 5x plus 4. Now 2x would remain as it is that is first element of the column C2 would remain as it is and the first column of the column C3 would also remain as it is. Then this 2x that is the second element of the column C1 is represented by 2x plus x plus 4 plus 2x and that would be 5x plus 4. Now the second element of the column C2 would remain as it is and the second element of the column C3 would remain as it is. Now the third element of the column C1 that is 2x is represented as 2x plus 2x plus x plus 4 and that is 5x plus 4. Then the third element of column C2 would remain as it is and third element of the column C3 would remain as it is. So, further we get this delta would be equal to, now we can take out 5x plus 4 common from the first column, so we have 5x plus 4 the whole multiplied by the determinant with elements in the first row as 1, 2x, 2x, but here it would be 1 since we have taken 5x plus common then in the second row the elements would be 1x plus 4, 2x in the third row the elements would be 1, 2x and x plus 4. Next applying the operations r2 goes to r2 minus r1 and r3 goes to r3 minus r1 we get delta is equal to 5x plus 4 the whole multiplied by the determinant obtained by applying these operations. Now the first row would remain as it is that is 1, 2x, 2x now applying the operation r2 goes to r2 minus r1 to the row 2 that is r2 we get the first element of the row r2 that is 1 is given by 1 minus the first element of the row r1 that is 1 and this would be 0. Now the second element of 0 r2 is given by x plus 4 minus 2x and it would be 4 minus x then the third element of r2 that is 2x is given by 2x minus 2x that is 0. Now let's find out the elements of the third row that is r3 which is represented by r3 minus r1 so the first element of the row r3 would be given by 1 minus 1 it would be 0 then the second element of r3 is 2x minus 2x that would be 0 and the third element of r3 would be x plus 4 minus 2x and that would be 4 minus x. Next we expand the given determinant along c1 so we get delta is equal to 5x plus 4 the whole multiplied by this 1 into 4 minus x into 4 minus x that is 4 minus xt whole square minus 0. Now the other two elements of the column c1 are 0 and 0 so here we would have minus 0 plus 0 and so further we get delta is equal to 5x plus 4 the whole multiplied by 4 minus xt whole square thus we get the given determinant is equal to 5x plus 4 the whole multiplied by 4 minus xt whole square. Hence proved this completes the session hope you have understood the solution of this question.