 Hello and welcome to the session. In this session, we discussed the following question which says, using matrices, solve the following system of equations. 2x minus 3y plus 5z is equal to 11, 3x plus 2y minus 4z equal to minus 5, x plus y minus 2z equal to minus 3. Let's move on to the solution now. The given system of equations given to us is 2x minus 3y plus 5z equal to 11. Let this be equation 1, 3x plus 2y minus 4z equal to minus 5. Let this be equation 2, x plus y minus 2z equal to minus 3. Let this be equation 3. We can write these three equations in the form, ax equal to b where we have a is equal to the matrix with elements in the first row as 2 minus 3, 5. In the second row, the elements are 3, 2, minus 4. In the third row, the elements are 1, 1, minus 2. And we take x equal to the column matrix with elements x, y, z. And b is equal to the column matrix with elements as 11, minus 5, minus 3. Now first of all, we will find the determinant of a. This is equal to determinant with elements in the first row as 2, minus 3, 5. In the second row, the elements are 3, 2, minus 4. In the third row, the elements are 1, 1, minus 2. This is equal to 2 multiplied by 2 multiplied by minus 2, that is minus 4, minus minus 4 into 1, that is minus 4. Then minus of minus 3 into 3 multiplied by minus 2, that is minus 6, minus 1 multiplied by minus 4, that is minus 4. Plus 5 multiplied by 3 multiplied by 1, that is 3, minus 1 multiplied by 2, that is 2. And so this is equal to 2 multiplied by minus 4, plus 4, that would be 0. Then plus 3, minus 6, plus 4 is minus 2, plus 5 into 3, minus 2 is 1. And so this is equal to minus 6, plus 5, which is equal to minus 1. Thus determinant of a is equal to minus 1, which is not equal to 0. Now as the determinant of a is not equal to 0, this means that a inverse exists and a inverse is equal to 1 upon determinant of a into a joint of a. Now determinant of a is equal to determinant with elements in the first row as 2 minus 3, 5. In the second row the elements are 3, 2, minus 4. In the third row the elements are 1, 1, minus 2. Now the adjoint of the matrix a is equal to the transpose of the matrix with elements in the first row as a11, a12, a13. In the second row the elements are a21, a22, a23. In the third row the elements are a31, a32, a33. That means this is equal to the matrix with elements in the first row as a11, a21, a31. In the second row a12, a22, a32. In the third row a13, a23, a33. So this is the adjoint of a. Now these are the cofactors of the elements of the determinant a. Now a11 is equal to the determinant with elements in the first row as 2 minus 4 and in the second row as 1 minus 2. This is equal to 2 into minus 2 minus minus 4 into 1. That is this is equal to minus 4 plus 4 which is equal to 0. This is a11. Now a12 is equal to minus of the determinant with elements in the first row as 3 minus 4 and in the second row as 1 minus 2. So this is equal to minus of 3 into minus 2 minus 1 into minus 4. So this is equal to minus of minus 6 plus 4. That is equal to plus 2. So we have a12 is equal to 2. Now a13 is equal to determinant with elements in the first row as 3, 2 and elements in the second row as 1, 1. So this is equal to determinant 3, 2, 1, 1 which is equal to 3 into 1 that is 3 minus 2 into 1 that is 2 and this is equal to 1. That is a13 is equal to 1. In the same way we can find out a21, a22, a23, a31, a32 and a33. Now a21 would be equal to minus 1. a22 would be equal to minus 9. a23 would be equal to minus 5. Then a31 would be equal to 2. a32 would be equal to 23. a33 would be equal to 13. So thus we get a joint of a is equal to matrix with elements in the first row as a11, a21 and a31. Now a11 is 0, a21 is minus 1 and a31 is 2. So elements in the first row of the matrix a joint of a would be 0 minus 1, 2. Elements of the second row of the adjoint of a would be a12, a22 and a32. a12 is 2, a22 is minus 9 and a32 is 23. So here we have 2 minus 9, 23. Then the elements of the third row of adjoint of a would be a13, a23 and a33. Now a13 is 1, a23 is minus 5 and a33 is 13. So here we have 1 minus 5, 13. a inverse is equal to 1 upon determinant of a into adjoint of a. Now determinant of a is minus 1. So this is equal to minus 1 into adjoint of a which is the matrix with elements in the first row as 0 minus 1, 2. In the second row the elements are 2 minus 9, 23. In the third row the elements are 1 minus 5, 13. So this is a inverse which is equal to matrix with elements in the first row as 0, 1, minus 2. In the second row minus 2, 9, minus 23. In the third row minus 1, 5, minus 13. This is a inverse. We have taken ax equal to b. Now since ax is equal to b, so from here we get x is equal to a inverse into b. That is x would be equal to a inverse. That is this matrix multiplied by the column matrix b with elements 11, minus 5, minus 3. So further we have column matrix with elements. x, y, z is equal to the matrix with elements. That is the column matrix with elements 0 into 11, that is 0, plus 1 into minus 5, that is minus 5, plus minus 2 into minus 3, that is 6. Then the second element of this column matrix is minus 2 into 11, that is minus 22, plus 9 into minus 5, which is minus 45, plus minus 23 into minus 3, that is 69. Then the third element is minus 1 into 11, which is minus 11, plus 5 into minus 5, minus 25, plus minus 13 into minus 3, that is 39. So further this would be equal to the column matrix with elements 1, 2, 3. That is the column matrix x, y, z is equal to the column matrix with elements 1, 2, 3. And from here we get x is equal to 1, y is equal to 2 and z is equal to 3. So this is our final answer. This completes the session. Hope you have understood the solution of this question.