Find inverse of A = 1 2 4 5 by elementary row transformation method

Let A = I × A 1 2 4 5 = 1 0 0 1 A 1 2 0 - 3 = 1 0 - 4 1 A , applying R 2 → R 2 - 4 R 1 1 2 0 1 = 1 0 4 3 - 1 3 A , applying R 2 → 1 - 3 R 2 1 0 0 1 = - 5 3 2 3 4 3 - 1 3 A , applying R 1 → R 1 - 2 R 2 Therefore, - 5 3 2 3 4 3 - 1 3 A = I ⇒ A - 1 = - 5 3 2 3 4 3 - 1 3

Find inverse of A = 1 2 4 5 by elementary row transformation method

Let A = I × A 1 2 4 5 = 1 0 0 1 A 1 2 0 - 3 = 1 0 - 4 1 A , applying R 2 → R 2 - 4 R 1 1 2 0 1 = 1 0 4 3 - 1 3 A , applying R 2 → 1 - 3 R 2 1 0 0 1 = - 5 3 2 3 4 3 - 1 3 A , applying R 1 → R 1 - 2 R 2 Therefore, - 5 3 2 3 4 3 - 1 3 A = I ⇒ A - 1 = - 5 3 2 3 4 3 - 1 3

Find the values of a , b , c and d from the following equations: 2 a + b a - 2 b 5 c - d 4 c + 3 d = 4 - 3 11 24 .

Given that, 2 a + b a - 2 b 5 c - d 4 c + 3 d = 4 - 3 11 24 To find : value of a , b , c and d Now, 2 a + b a - 2 b 5 c - d 4 c + 3 d = 4 - 3 11 24 Equating the corresponding elements... 2 a + b = 4 ⇒ b = 4 - 2 a . . . . . . . . . . . 1 a - 2 b = - 3 . . . . . . . . . . . . . 2 ⇒ a - 2 4 - 2 a = - 3 [from (1)] ⇒ a - 8 + 4 a = - 3 ⇒ a = 1 Put the value of a in equation 1 b = 4 - 2 1 ⇒ b = 2 5 c - d = 11 ⇒ 5 c - 11 = d . . . . . . . . . . 3 4 c + 3 d = 24 . . . . . . . . . . . 4 ⇒ 4 c + 3 5 c - 11 = 24 [from (3)] ⇒ c = 3 Put the value of c in 3 5 × 3 - 11 = d ⇒ d = 4 . Hence, a = 1 , b = 2 , c = 3 , d = 4 .

If A = 8 0 4 - 2 3 6 and B = 2 - 2 4 2 - 5 1 , then find the matrix X , such that 2 A + 3 X = 5 B .

A = 8 0 4 - 2 3 6 and B = 2 - 2 4 2 - 5 1 Since, 2 A + 3 X = 5 B ⇒ 2 8 0 4 - 2 3 6 + 3 X = 5 2 - 2 4 2 - 5 1 ⇒ 16 0 8 - 4 6 12 + 3 X = 10 - 10 20 10 - 25 5 ⇒ 3 X = 10 - 10 20 10 - 25 5 - 16 0 8 - 4 6 12 ⇒ X = 1 3 - 6 - 10 12 14 - 31 - 7

Find A B if A = 6 9 2 3 and B = 2 6 0 7 9 8 .

Given, A = 6 9 2 3 and B = 2 6 0 7 9 8 AB is defined as the number of columns in matrix A is same as the number of rows in matrix B . Also AB is of order 2 × 3 . AB = 6 9 2 3 2 6 0 7 9 8 Applying row and column multiplication... AB = 6 × 2 + 9 × 7 6 × 6 + 9 × 9 6 × 0 + 9 × 8 2 × 2 + 3 × 7 2 × 6 + 3 × 9 2 × 0 + 3 × 8 AB = 12 + 63 36 + 81 0 + 72 4 + 21 12 + 27 0 + 24 AB = 75 117 72 25 39 24

If X + Y = 4 3 2 5 and X - Y = 4 0 0 4 . Find X and Y .

Given: X + Y = 4 3 2 5 and X - Y = 4 0 0 4 Adding the above two we get, X + Y + X - Y = 4 3 2 5 + 4 0 0 4 ⇒ 2 X = 8 3 2 9 ⇒ X = 1 2 8 3 2 9 = 4 1 . 5 1 4 . 5 Putting X in first equation we get, 4 1 . 5 1 4 . 5 + Y = 4 3 2 5 ⇒ Y = 4 3 2 5 - 4 1 . 5 1 4 . 5 ⇒ Y = 0 1 . 5 1 0 . 5

If A = 3 2 4 5 . Find A 2

Given: A = 3 2 4 5 . The matrix A can be multiplied with itself as the condition for multiplication, that is, number of columns in pre multiplier must be equal to number of rows in the post multiplier, is being satisfied. A 2 = 3 2 4 5 × 3 2 4 5 = 9 + 8 6 + 10 12 + 20 8 + 25 , multiplying the rows of pre multiplier with the columns of post multiplier element wise. ∴ A 2 = 17 16 32 33

If A = 2 4 3 2 , B = 1 3 - 2 5 , C = - 2 5 3 4 . Find the following (i) 3 A - C (ii) 2 A - B - 3 C

Given: A = 2 4 3 2 , B = 1 3 - 2 5 , C = - 2 5 3 4 i 3 A - C = 3 2 4 3 2 - - 2 5 3 4 = 6 12 9 6 - - 2 5 3 4 = 8 7 6 2 i i 2 A - B - 3 C = 2 2 4 3 2 - 1 3 - 2 5 - 3 - 2 5 3 4 = 4 8 6 4 - 1 3 - 2 5 - - 6 15 9 12 = 4 - 1 + 6 8 - 3 - 15 6 + 2 - 9 4 - 5 - 12 = 9 - 10 - 1 - 13

If A = 0 0 1 0 1 0 1 0 0 , verify that A 2 = I .

Given: A = 0 0 1 0 1 0 1 0 0 A 2 = 0 0 1 0 1 0 1 0 0 × 0 0 1 0 1 0 1 0 0 = 0 + 0 + 1 0 + 0 + 0 0 + 0 + 0 0 + 0 + 0 0 + 1 + 0 0 + 0 + 0 0 + 0 + 0 0 + 0 + 0 0 + 0 + 1 applying the multiplication of matrices taking rows from the pre multiplier and columns from the post multiplier. ⇒ A 2 = 1 0 0 0 1 0 0 0 1 = I where I is the unit matrix.