Find the inverse of the following matrix. 2 - 3 - 1 2

Given that, 2 - 3 - 1 2 To find : inverse matrix of given matrix. Now, Applying row operations, consider A = I A 2 - 3 - 1 2 = 1 0 0 1 A R 1 → 1 2 R 1 1 - 3 2 - 1 2 = 1 2 0 0 1 A R 2 → R 2 - - 1 R 1 1 - 3 2 0 1 2 = 1 2 0 1 2 1 A R 2 → 2 R 2 1 - 3 2 0 1 = 1 2 0 1 2 A R 1 → R 1 - - 3 2 R 2 1 0 0 1 = 2 3 1 2 A Hence, A - 1 = 2 3 1 2 .

Find the inverse of the following matrix. 2 - 3 - 1 2

Given that, 2 - 3 - 1 2 To find : inverse matrix of given matrix. Now, Applying row operations, consider A = I A 2 - 3 - 1 2 = 1 0 0 1 A R 1 → 1 2 R 1 1 - 3 2 - 1 2 = 1 2 0 0 1 A R 2 → R 2 - - 1 R 1 1 - 3 2 0 1 2 = 1 2 0 1 2 1 A R 2 → 2 R 2 1 - 3 2 0 1 = 1 2 0 1 2 A R 1 → R 1 - - 3 2 R 2 1 0 0 1 = 2 3 1 2 A Hence, A - 1 = 2 3 1 2 .

Find the inverse of the following matrix. 0 1 2 1 2 3 3 1 1

Given matrix is A = 0 1 2 1 2 3 3 1 1 Now, A = I A ⇒ 0 1 2 1 2 3 3 1 1 = 1 0 0 0 1 0 0 0 1 A Applying R 1 → R 1 + R 2 , R 3 → R 3 - 3 R 2 , then we get: 1 3 5 1 2 3 0 - 5 - 8 = 1 1 0 0 1 0 0 - 3 1 A Applying R 2 → R 2 - R 1 , then we get: 1 3 5 0 - 1 - 2 0 - 5 - 8 = 1 1 0 - 1 0 0 0 - 3 1 A Applying R 1 → R 1 + 3 R 2 , R 3 → R 3 - 5 R 2 , then we get: 1 0 - 1 0 - 1 - 2 0 0 2 = - 2 1 0 - 1 0 0 5 - 3 1 A Applying R 3 → 1 2 R 3 , R 2 → ( - 1 ) R 2 , then we get: 1 0 - 1 0 1 2 0 0 1 = - 2 1 0 1 0 0 5 2 - 3 2 1 2 A Applying R 1 → R 1 + R 3 , R 2 → R 2 - 2 R 3 , then we get: 1 0 0 0 1 0 0 0 1 = 1 2 - 1 2 1 2 - 4 3 - 1 5 2 - 3 2 1 2 A ⇒ I = A - 1 · A Hence, the inverse of the given matrix is A - 1 = 1 2 - 1 2 1 2 - 4 3 - 1 5 2 - 3 2 1 2 .

Find the inverse of the following matrix. 2 0 - 1 5 1 0 0 1 3

We have, A = I A or 2 0 − 1 5 1 0 0 1 3 = 1 0 0 0 1 0 0 0 1 A Applying R 1 → R 1 2 ⇒ 1 0 − 1 2 5 1 0 0 1 3 = 1 2 0 0 0 1 0 0 0 1 A Applying R 2 → R 2 - 5 R 1 ⇒ 1 0 − 1 2 0 1 5 2 0 1 3 = 1 2 0 0 - 5 2 1 0 0 0 1 A Applying R 3 → R 3 - R 2 ⇒ 1 0 − 1 2 0 1 5 2 0 0 1 2 = 1 2 0 0 - 5 2 1 0 5 2 - 1 1 A Applying R 3 → 2 R 3 ⇒ 1 0 − 1 2 0 1 5 2 0 0 1 = 1 2 0 0 - 5 2 1 0 5 - 2 2 A Applying R 1 → R 1 + 1 2 R 3 ⇒ 1 0 0 0 1 5 2 0 0 1 = 3 - 1 1 - 5 2 1 0 5 - 2 2 A Applying R 2 → R 2 - 5 2 R 3 ⇒ 1 0 0 0 1 0 0 0 1 = 3 - 1 1 - 15 6 - 5 5 - 2 2 A Hence, A - 1 = 3 − 1 1 − 15 6 − 5 5 − 2 2

If A = 1 0 0 2 1 0 3 3 1 then reduce it to I 3 by using column transformations.

| A | = 1 ( 1 - 0 ) - 0 ( 2 - 0 ) + 0 ( 6 - 3 ) = 1 ∴ A ≠ 0 ⇒ A - 1 exists So, we can reduce A to I 3 Consider A = 1 0 0 2 1 0 3 3 1 Applying C 1 → C 1 - 2 C 2 , we get = 1 0 0 0 1 0 - 3 3 1 Applying C 1 → C 1 + 3 C 3 = 1 0 0 0 1 0 0 3 1 Applying C 2 → C 2 - 3 C 3 , we get = 1 0 0 0 1 0 0 0 1 = I 3 In this way, we reduced the given matrix A to I 3 using column transformation.

If A = 2 1 3 1 0 1 1 1 1 then reduce it to I 3 by using row transformations.

We have | A | = 1 ⇒ A - 1 exists. So, we can reduce A to I 3 using row transformations. Consider A = 2 1 3 1 0 1 1 1 1 Applying R 1 ↔ R 2 , we get = 1 0 1 2 1 3 1 1 1 Applying R 2 → R 2 - 2 R 1 , we get = 1 0 1 0 1 1 1 1 1 Applying R 3 → R 3 - R 1 , we get = 1 0 1 0 1 1 0 1 0 Applying R 3 ↔ R 2 , we get = 1 0 1 0 1 0 0 1 1 Applying R 3 → R 3 - R 2 , we get = 1 0 1 0 1 0 0 0 1 Applying R 1 → R 1 - R 3 , we get = 1 0 0 0 1 0 0 0 1 = I 3

Express the following equation in matrix form and solve them by the method of reduction x + y + z = 6 , 3 x - y + 3 z = 10 and 5 x + 5 y - 4 z = 3

Given equations are x + y + z = 6 , 3 x - y + 3 z = 10 and 5 x + 5 y - 4 z = 3 The above equations can be written in the form A X = B as 1 1 1 3 - 1 3 5 5 - 4 x y z = 6 10 3 Using R 2 → R 2 - 3 R 1 , we get 1 1 1 0 - 4 0 5 5 - 4 x y z = 6 - 8 3 Using R 3 → R 3 - 5 R 1 , we get 1 1 1 0 - 4 0 0 0 - 9 x y z = 6 - 8 - 27 we have reduced the original matrix A to a upper triangular matrix. By equality of matrices x + y + z = 6 ...(1) - 4 y = - 8 ⇒ y = 2 ....(2) - 9 z = - 27 ⇒ z = 3 ....(3) substituting y = 2 , z = 3 in (1), we get x + 2 + 3 = 6 ⇒ x = 1 Thus, x = 1 , y = 2 , z = 3 is the required solution