 Hello and welcome to the session. Let us discuss the following question. It says using matrices solve the following system of equation. So let's now move on to the solution. The given system of equation is 3x minus y plus z is equal to 5 2x minus 2y plus 3z is equal to 7 x plus y minus z is equal to minus 1. Now the system of equation can be written as ax is equal to b. So this implies x is equal to a inverse b where a is the matrix where first row consists of the coefficient of x, y and z of the first equation that is 3 minus 1, 1. Second row consists of the coefficient of x, y and z of the second equation. There's 2 minus 2 3 and similarly the third row 1, 1 minus 1 and x is the matrix consisting of the unknown variables x, y, z. So call a matrix and b is the column matrix consisting of the elements on the right hand side and these are 5, 7, minus 1. Now to solve for x, y and z we need to find a inverse. Now a inverse is given by 1 upon the determinant of a into a joint of a. Now we find the determinant of a 3 minus 1, 1, 2, minus 2, 3, 1, 1, minus 1. Let's now expand this determinant along first row. So we have 3 into minus 1 into minus 2 is 2 minus 1 into 3 is 3 minus of minus 1 is plus 1 into minus 2 1 into 3 is 3 plus 1 into 2 into 1 is 1 minus of minus 2 is plus 2 so this is equal to 3 into minus 1 plus 1 into minus 5 plus 1 into 4 this is equal to minus 3 minus 5 plus 4. Now minus 3 minus 5 is minus 8 plus 4 is minus 4 and this is not equal to 0. So this implies inverse of a exists. Now we have to find the adjoint of a or the adjoint of a is given by the matrix a 1, 1, a 2, 1, a 3, 1 a 1, 2, a 2, 2, a 3, 2, a 1, 3, a 2, 3, a 3, 3 where aij is minus 1 to the power i plus j into mij where mij is the minor obtained by deleting ith row and jth column in which the element of the matrix aij lies now we find a 1, 1, a 1, 1 is minus 1 to the power 1 plus 1 that is 2 into m 1, 1, m 1, 1 is the minor obtained by deleting first row and the first column so it is minus 1 into minus 2 that is 2 minus 1 into 3 that is 2 minus 3 so this is equal to minus 1 to the power 2 into 2 minus 3 that is minus 1 right similarly we will find a 2, 1, a 2, 1 is minus 1 to the power 2 plus 1 that is 3 into m 2, 1, m 2, 1 is the minor obtained by deleting second row and first column so it is minus 1 into minus 1 is 1 minus 1 into 1 that is minus 1 to the power 3 into 1 minus 1 that is 0 so in the similar manner we have obtained all the cofactors now adjoint a minus 1, 0 minus 1, 5 minus 4 minus 7, 4 minus 4 minus 4 now x is a inverse b now a inverse is 1 upon determinant of a now determinant of a is minus 1 upon determinant of a is minus 4 so here we have 1 upon minus 4 into adjoint of a which is minus 1, 0 minus 1, 5 minus 4 minus 7, 4 minus 4, minus 4 into b, b is the column matrix 5, 7, minus 1 now we will multiply the two matrices let's now multiply the two matrices we have minus 1 by 4 into minus 1 into 5 is minus 1 plus 0 into 7 is 0 plus minus 1 into minus 1 is plus 1, 5 into 5 is 25 minus 4 into 7 is minus 28 minus 7 into minus 1 is plus 7 minus 7 into minus 1 is plus 7, 4 into 5 is 20 minus 4 into 7 is minus 28 minus 4 into minus 1 is plus 4 this is equal to minus 1 by 4 into minus 5 plus 1 is minus 4 25 minus 28 plus 7 is 4 and 20 minus 28 plus 4 is minus 4 dividing each element by minus 4 that is minus 4 by minus 4, 4 divided by minus 4 minus 4 by minus 4 so this is equal to 1 minus 1, 1 so the matrix x is equal to x, y, z which is equal to 1 minus 1, 1 so we have x is equal to 1, y is equal to minus 1 and z is equal to 1 is the solution of the given system of equations so this completes the question and the session bye for now take care have a good day