 Hello and welcome to the session. The given question says using matrix method solve the following system of equations So let's begin with the solution. Here we are given three equations. First one is x plus 2y plus z is equal to 7 Second is x plus 3z is equal to 11 and the third one is 2x Minus 3y is equal to 1 Now we have to solve these three equations using matrix method That is we have to find the values of x, y and z which satisfies these given equations So writing these three equations in matrix form, we have on the left hand side 1 2 1 1 0 3 2 minus 3 0 and we have x, y, z is equal to 7, 11 and 1 Let us denote this matrix by a, this by x and this by b So we have a x is equal to b, we have to find x and this is equal to a inverse of b and a inverse exists only if a is a non-singular matrix. So first let us find determinant of a So this is equal to 1 into 0 minus of minus 9 minus 2 times of 0 minus of 6 Plus 1 into minus 3 minus 0. So this is equal to 9 Plus 12 minus 3 which is equal to 18 So determinant of a is equal to 18 and this is not equal to 0 So this implies that a inverse exists and a inverse is given by a Joint of a divided by determinant of a Now let us find a joint of a now a is a matrix having elements 1 2 1 1 0 3 and 2 Minus 3 0 So here we shall write the cofactor of a 1 1 which is given by plus remaining After leaving the first row and the first column we are left with the remaining these four elements which is the cofactor of a 1 1 and This gives 0 plus 9 which is equal to 9 So writing 9 in the place of a 1 1 now let us find a 2 1 But just this element so leaving this row and this column we are left with 2 1 minus 3 0 and the sum of 2 and 1 is minus 3 So here we shall be writing minus sign and this gives minus of plus 3 which is equal to minus 3 so writing this in The place of a 1 2 Now let us write a 3 1 Thus 3 plus 1 is 4 which is a positive number. So here we have 4 and a 3 1 is this element So leaving the third row and the first column we have 2 1 0 3 and this gives 6 so here we have 6 now let us find a 1 2 Again we have a negative sign and a 1 2 is this element. So leaving this column and the first row we have 1 2 3 0 and this is equal to Minus of minus 6 which is equal to 6. So here we have 6 similarly writing the others We have this So this is a joint of a now. Let us find a inverse. So this is equal to a Joint of a which is this matrix 9 minus 3 6 6 minus 2 minus 2 minus 3 7 and minus 2 divided by determinant of a which is 18 and Thus x is equal to a inverse which is this matrix that is 1 upon 18 9 minus 3 6 6 minus 2 minus 2 minus 3 7 minus 2 and B Which is a column matrix having elements 7 11 and 1 and this is further equal to 1 divided by 18 Now 9 7's are 63 3 into 11 is minus 33 6 ones are 6 Then we have 6 7's are 42 2 and 2 11 gives 22 with a negative sign minus 22 minus 2 into 1 is minus 2 and here we have 3 into 7 with a negative sign minus 21 7 into 11 is 77 and minus 2 into 1 is minus 2 This further implies That 1 divided by 18 On simplifying this matrix we get the first 36 Then we get 18 and I'm simplifying the last Row we have 54 Which further implies that 36 divided by 18 18 divided by 18 and 54 divided by 18 Which is equal to 2 1 and 3 Therefore x whose elements are small x small y and small z are 2 1 and 3 So on comparing we have x is equal to 2 y is equal to 1 and Z is equal to 3 so this completes the session by intake care