 Hi, and welcome to the session. Here is discuss the following question. The question says, use matrices. Solve the following system of equations. x plus 5 plus z is equal to 3, x minus 2 y plus 3 z is equal to 2, and 2x minus 5 plus z is equal to 2. Let's now begin with the solution. Given equations are x plus 5 plus z is equal to 3, x minus 2 y plus 3 z equals to 2, and 2x minus 5 plus z is equal to 2. Now these given equations can be written as equals to b, where a is equal to 3 by 3 matrix in which elements are coefficient of x, y, and z. This is equal to column matrix in which elements are x, y, and z. b is also a column matrix in which elements are 3, 2, 2. Find determinant of a, 1 into determinant of minus 2, 3, minus 1, 1, minus 1 into determinant of 1, 2, 3, 1, plus 1 into determinant of 1, 2, minus 2, minus 1. Now this is equal to 1 into minus 2 plus 3, minus 1 into 1 minus 6, plus 1 into minus 1 plus 4. This is equal to 1 into 1, minus 1 into 5, sorry, minus 5, plus 1 into 3. This is equal to 1 plus 5 plus 3. And this is equal to 9, which is not equal to 0. Determinant of a is not equal to 0, so this implies given system of equations unique solution and system of equations by Grammler's rule. We'll first find d1. d1 is the determinant of elements given on the right-hand side, coefficients of z, and coefficients of 5. So d1 is equal to determinant of minus 1, 1, 3, 1. 3 into minus 2, plus 3, minus 1, into 2, minus 6, plus 1, into minus 2, plus 4. This is equal to 3 into 1, minus 1, into minus 4, plus 1, into 2. This is equal to 3, plus 4, plus 2, and this is equal to d1 is equal to 9. Now we will find determinant of d2. Coefficient of x is equal to 1, into 2, minus 1, into 2, minus 4. And this is equal to minus 4, plus 15, minus 2, and this is equal to 9. Terminant of d3. In this determinant, elements will be coefficient of y is equal to minus 1, 3 is equal to 1, into minus 4, plus 2, minus 1, into 2, minus 4, plus 3, into minus 1, plus 1, into minus 2, plus 3, into 3, plus 2, minus 2, and this is equal to determinant of d1, by determinant of a, y is equal to determinant of d2, by determinant of a, and z is equal to determinant of d3, by determinant of a. Now determinant of d1 is 9, and determinant of a is also 1, and 9 by 9 is equal to 1. Determinant of d2 is also equal to 9, and determinant of a is also equal to 9. So y is also equal to 1. Determinant of d3 is equal to 9, and determinant of a is also equal to 9, 9 by 9 is equal to 1. So it equals to 1, y equals to 1, and z equals to 1 is the required solution of the given equation. So this completes the session. Bye and take care.