Using elementary row operations, show that the inverse of (2 5 8 5 1 2 3 1 2 4 7

Question

Using elementary row operations, show that the inverse of (2 5 8 5 1 2 3 1 2 4 7 2 1 3 5 3) is (3 -2 1 -5 -2 5 -2 3 0 -2 1 0 1 -1 0 -1) b) Without performing any further elementary row operations, use part (a) to solve the system of linear equations 2x_1 + 5x_2 + 8x_3 + 5x_4 = 0, x_1 + 2x_2 + 3x_3 + x_4 = 1, 2x_1 + 4x_2 + 7x_3 + 2x_4 = 0, x_1 + 3x_2 + 5x_3 + 3x_4 = 1. Consider the matrix A = (1 0 3 1 1 1 5 5 2 1 9 8 2 0 6 3) a) Use elementary row operations to find the inverse of A. b) Without performing any further elementary row operations, use your solution in part (a) to solve the system of linear equations x_1 + 3x_3 + x_4 = 1, x_1 + x_2 + 5x_3 + 5x_4 = 0, 2x_1 + x_2 + 9x_3 + 8x_4 = 0, 2x_1 + 6x_3 + 3x_4 = 0.

Explanation / Answer

Here is the edited and detailed solution for the previous answer as you asked-

Solution 7 part (a)

Write the augmented matrix

What we basically wants is to convert our left side 4*4 matrix into correspondig unity matrix and our answer will be the chages it automatically did to our right side unity matrix.

Here is the details of operations performed to convert our left side

4*4 matrix into unity matrix.

Find the pivot in the 1st column and swap the 2nd and the 1st rows

Multiply the 1st row by 2

Subtract the 1st row from the 2nd row2

Subtract the 1st row from the 3rd row and restore it

Subtract the 1st row from the 4th row4

Find the pivot in the 2nd column in the 2nd row

Multiply the 2nd row by 2

Subtract the 2nd row from the 1st row and restore it

Subtract the 2nd row from the 4th row4

Find the pivot in the 3rd column in the 3rd row

Multiply the 3rd row by -1

Subtract the 3rd row from the 1st row and restore it

Multiply the 3rd row by 2

Subtract the 3rd row from the 2nd row and restore it

Find the pivot in the 4th column in the 4th row (inversing the sign in the whole row)

Multiply the 4th row by -5

Subtract the 4th row from the 1st row and restore it

Multiply the 4th row by 3

Subtract the 4th row from the 2nd row and restore it

There is the inverse matrix on the right

Result:

Solution part 7(b)

C = A-1B

thus We already have inverse of A

and B=

1

c11 = 3 x 0 + (-2) x 1 + 1 x 0 + (-5) x 1 = -7

c21 = -2 x 0 + 5 x 1 + (-2) x 0 + 3 x 1 = 8

c31 = 0 x 0 + (-2) x 1 + 1 x 0 + 0 x 1 = -2

c41 = 1 x 0 + (-1) x 1 + 0 x 0 + (-1) x 1 = -2

Result:

thus solution of equation is

x1= -7

x2 = 8

x3 = -2

x4 = -2

You can follow the same steps for question 8 too.

Comment and let me know if you require solution of question 8 too.

A1 A2 A3 A4 A1 A2 A3 A4 1 2 5 8 5 1 0 0 0 2 1 2 3 1 0 1 0 0 3 2 4 7 2 0 0 1 0 4 1 3 5 3 0 0 0 1

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