 Hi and welcome to the session. I'm Shashi and I'm going to help you with the following question. Question says, using matrices solve the following system of equations. First of all, let us understand that unique solution of the equation AX is equal to B is given by X is equal to A inverse B where determinant of A is not equal to 0. This is the key idea behind the given question. Let us now start with the solution. Now given equations are X plus 2Y minus 3Z is equal to 6. Second equation is 3X plus 2Y minus 2Z is equal to 3. And third equation is 2X minus Y plus Z is equal to 2. Now these equations can be written as AX is equal to B where A is equal to matrix 1, 2 minus 3, 3, 2 minus 2, 2 minus 1, 1. And X is equal to matrix XYZ also B represents the matrix 6, 3, 2. Now first of all we will find out determinant of A. It is given by determinant of 1, 2 minus 3, 3, 2 minus 2, 2 minus 1, 1. Now expanding this determinant with respect to first row we get determinant of A is equal to 1 multiplied by 2 minus 2 minus 2 multiplied by 3 minus minus 4 plus minus 3 multiplied by minus 3. Now this is further equal to minus 14 plus 21. Clearly we can see this term will become 0, 2 minus 2 is 0 and 0 multiplied by 1 is equal to 0. 3 minus minus 4 is equal to 3 plus 4 that is 7 and minus 2 multiplied by 7 is equal to minus 14. Now minus 3 minus 4 is equal to minus 7 and minus 7 multiplied by minus 3 is equal to plus 21 and minus 14 plus 21 is equal to 7. Now clearly we can see determinant of A is equal to 7 implies that determinant of A is not equal to 0. So given system of equations have a unique solution or we can say A is a non-singular matrix so its inverse exists. Now we know inverse of A is equal to adjoint A upon determinant of A. Now to find adjoint A we will find the cofactors of the elements of the determinant A. Clearly we can see this is the determinant A. Now cofactor A11 is equal to minus 1 raised to the power 1 plus 1 multiplied by minus 2 minus 2 minus 1 1. Now this is further equal to below minus 1 raised to the power 2 is equal to 1 only. So here we can write 1 and solving this determinant we get 2 minus 2 that is 0. So A11 is equal to 0. Now similarly we can find cofactor A12 it is equal to minus 1 raised to the power 1 plus 2 multiplied by minus 3 minus 2 2 1. Now this is further equal to below 1 plus 2 is equal to 3 and minus 1 raised to the power 3 is equal to minus 1 only. And solving this determinant we get 3 minus minus 4 which is further equal to minus 1 multiplied by you know minus multiplied by minus sign is equal to plus. So we will get 3 plus 4 and here we can write 7 below 3 plus 4 is equal to 7. Now minus 1 multiplied by 7 is equal to minus 7. So cofactor A12 is equal to minus 7. Now we will find cofactor A13 it is equal to minus 1 raised to the power 1 plus 3 multiplied by minus 3 2 2 minus 1. Now we know 1 plus 3 is equal to 4 and minus 1 raised to the power 4 is equal to 1 only. So here we can write 1. Now we will solve this determinant minus 1 multiplied by 3 is equal to minus 3 and 2 multiplied by 2 is equal to 4. So here we will write minus 3 minus 4. Now 1 multiplied by minus 7 you know minus 3 minus 4 is equal to minus 7 only. So we get minus 7 is equal to cofactor A13. Now similarly we can find other cofactors also. A21 is equal to minus 1 raised to the power 2 plus 1 multiplied by minus 2 minus 3 minus 1 1. Now again solving it further we get 1. Now we will find cofactor A22 it is equal to minus 1 raised to the power 2 plus 2 multiplied by minus 1 minus 3 2 1. Now solving it further we get A22 is equal to 7. Now we will find A23 it is equal to minus 1 raised to the power 2 plus 3 multiplied by minus 1 2 2 minus 1. Now solving it further we get 5. Now similarly we can find other cofactors also A31 is equal to 2, A32 is equal to minus 7 and A33 is equal to minus 4. Now we know this matrix represents the adjoint of A. All the elements of this matrix are the cofactors of determinant A. Now substituting the corresponding values of all these cofactors in this matrix we get adjoint of A is equal to matrix 012 minus 7 7 minus 7 minus 7 5 minus 4. Now we also know that A inverse is equal to adjoint A upon determinant of A. Now we know determinant of A is equal to 7 and adjoint A is equal to this matrix. So we can write A inverse is equal to 1 upon 7 multiplied by matrix 012 minus 7 7 minus 7 minus 7 5 minus 4. Now from key idea we know x is equal to A inverse B. Now we know this matrix represents x and this matrix represents B. So we can write matrix xyz is equal to 1 upon 7 multiplied by matrix 012 minus 7 7 minus 7 minus 7 5 minus 4 multiplied by matrix 632. Now here in this expression we have replaced x by this matrix A inverse is equal to 1 upon 7 multiplied by this matrix and B represents this matrix. Now we will multiply these two matrices. We will write 1 upon 7 as it is. Multiplying this row by this column we get 0 plus 3 plus 4. Now multiplying this row by this column we get minus 42 plus 21 minus 14. Now multiplying this row by this column we get minus 42 plus 15 minus 8. Now this is further equal to 1 upon 7 multiplied by 7 adding these three terms we get 7. We know minus 42 plus 21 minus 14 is equal to minus 35. Similarly minus 42 plus 15 minus 8 is equal to minus 35. Now this is further equal to matrix 1 upon 7 multiplied by 7 is equal to 1. 1 upon 7 multiplied by minus 35 is equal to minus 5. Similarly here we will write minus 5 1 upon 7 multiplied by minus 35 is equal to minus 5 only. So we get matrix x y z is equal to matrix 1 minus 5 minus 5. So x is equal to 1 y is equal to minus 5 and z is equal to minus 5. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.