 Hello and welcome to the session. I'm Ashima and I'm there to help you with the following problem. Solving equation X plus AXxxxxxxxxxx. is, equal to 0. where A is not equal to 0. so we have this determinant, now let us write the solution. In the given problem, we have to find the value of x. We have determinant x plus a, x x, x, x plus a, x, x, x, x plus a. Which is equal to zero, which implies now applying r1 tens to r1 plus r2 plus r3. Now we are applying the row operation on row 1, so we get 3x plus a, 3x plus a, 3x plus a, xx plus a, xx, xx plus a is equal to 0. Now taking 3x plus a common from row 1 which implies 3x plus a multiplied by determinant 1, 1, 1, xx plus a, xx, xx plus a which is equal to 0. Now applying C1 tends to C1 minus C3, now we will apply the column operation on column 1. So we get 3x plus a multiplied by determinant 0, 1, 1, 0, x plus a minus x, x minus x is 0, x minus x plus a is equal to minus a and these terms are written equal to 0 which implies 3x plus a. Now with spanning we get, we will expand it along minus a, so we get minus a multiplied by eliminating this row and this column we get x minus x plus a equal to 0 which implies 3x plus a multiplied by minus a multiplied by x minus x minus a which is equal to 0. Here this x gets cancelled with this x, so we get 3x plus a minus a into minus a which is equal to 0 which implies 3x plus a multiplied by a square which is equal to 0. Now it is given to us in the question that a is not equal to 0, now we see that here, it implies 3x plus a is equal to 0 or a square is equal to 0 that is a is equal to 0 but a is not equal to 0 is given in the question. So therefore 3x plus a is equal to 0 which implies 3x is equal to minus a and this implies x is equal to minus a divided by 3 therefore the required value of x is minus a by 3. I hope you understood the problem, bye and have a nice day.