 Hi and welcome to the session. I am Dipika here. Let's discuss the questions. All system of linear equations using matrix method. 2x plus 3y plus 3z is equal to 5, x minus 2y plus z is equal to minus 4, 3x minus 5, minus 2z is equal to 3. Let's start the solution. The given system of linear equation can be written in the form equal to b where equal to minus 2, 1, minus 1, minus 2, equal to x, y, z. b is equal to we have determinant a as 1 into minus 2, minus 2, minus 1 into 3. Again, plus 3 into 1 into minus 1, minus 3 into minus 2, which is equal to plus 6. So, this is equal to minus 3 into minus 5 plus 3 into 5. This is equal to 10 plus 15 plus 15 and it is equal to 40 which is not equal to 0. This implies, since determinant a is not equal to 0, a is non-singular and so its inverse exists. Now, we will find out a inverse. For a inverse, let us first find out a joint a and for a joint a, we will find the co-factor of each and every element of a. Now, we know that the co-factor of 2 is equal to minus 1 raise to power 1 plus 1 into 4 plus 1. Again, co-factor of 3 is delete this column and this row which is equal to minus 1 raise to power 1 plus 2 into minus 2 minus 3. This is equal to 5. So, by this way we will find the co-factor of each and every element of a. This is as follows. Now, we have found all co-factors of all the elements. Therefore, the matrix formed by the co-factors is 5, 5, 3, minus 13, 11, 9, 1, minus 7. A joint a is equal to transpose of this matrix which is equal to 5, 5, 5. This is a matrix we have found by the co-factors. Now, take its transpose which is equal to 5, 5, 5, 3, minus 13, 11, 9, 1, minus 7. Therefore, a inverse is equal to 1 over determinant a into adjoint a which is equal to 1 over 40 into 5, 5, 5, 3, minus 13, 11, 9, 1, minus 7. Our system of linear equation in the form ax is equal to b which implies a inverse ax is equal to a inverse b pre-multiplying both sides by a inverse. So, this implies ax is equal to a inverse b which implies ax is equal to a inverse b as a inverse a is equal to i. That is identity matrix. Therefore, ax is equal to 1 over 40 into 5, 5, 5, minus 13, 11, 9, 1, minus 7, a inverse into b. Our b is 5, minus 4, 3. Now, we will solve. This is equal to 1 over 40 into 5, 5, 0, 25, minus 12, plus 27, 5, 5, 0, 25, plus 52, plus 1 into 3, 3, 5, 5, 0, 25, minus 44, minus 21. This is equal to 1 over 40 into, on solving here we get 40, here 80 and this is minus 40. This is equal to 1, 2, minus 1. This implies y, z is equal to 1, 2, minus 1. Now equating the corresponding elements, we get x is equal to 1, y is equal to 2 and z is equal to minus 1. Hence, we have solved the above system of linear equations using the matrix method and our answer is x is equal to 1, y is equal to 2, z is equal to minus 1. I hope the question is clear to you. Bye and have a good day.