Given the matrix A 1 2 2 5 5 7 1 0 1 Show your work to do t

Given the matrix A = [1 2 2 5 5 7 1 0 1] Show your work to do the following: a) Use the Elementary Row Operation to solve for A^-1. b) Use the A^-1 = 1/det(A) adj(A) to solve for A^-1. Given the system of equation { x_1 + 3x_2 + 2x_3 = 2 2x_1 + 7x_2 + 7x_3 = -1. 2x_1 + 5x_2 + 2x_3 = 7 Show your work to do the following: a) Use x = A^-1b to solve the linear system. b) Use the determinants to solve the linear system.

Solution

4. (a) Let B =

1

5

1

1

0

0

2

5

0

0

1

0

2

7

1

0

0

1

In order to determine A^-1 through row operations, we will reduce A to its RREF as under:

Add -2 times the 1st row to the 2nd row

Add -2 times the 1st row to the 3rd row

Multiply the 2nd row by -1/5

Add 3 times the 2nd row to the 3rd row

Multiply the 3rd row by 5

Add -2/5 times the 3rd row to the 2nd row

Add -1 times the 3rd row to the 1st row

Add -5 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

-5

-2

5

0

1

0

2

1

-2

0

0

1

-4

-3

5

Hence A^-1=

-5

-2

5

2

1

-2

-4

-3

5

(b) det(A) = 1(5*1-0*7)-5(2*1-0*2)+1(2*7-5*2) = 5-10+4 = -1.

To determine Adj(A), we will first compute the cofactors of the various entries in A. On starting from the left with the 1^st row,the cofactor of 1 is (5*1-0*7)= 5, the cofactor of 5 is –(2*1-0*2) =-2, and the cofactor of 1 is (2*7-5*2) = 4.

In the 2^nd row, the cofactor of 2 is –(5*1-1*7) = 2,the cofactor of 5 is (1*1-1*2) = -1, and the cofactor of 0 is -(1*7- 5*2) = 3.

In the 3^rd row, the cofactor of 2 is (5*0-1*5) = -5, the cofactor of 7 is –(1*0-1*2)=2, and the cofactor of 1 is (1*5-5*2)= -5.

Then the cofactor matrix of A is C =

5

-2

4

2

-1

3

-5

2

-5

Now, adj(A) = C^T =

5

2

-5

-2

-1

2

4

3

-5

Then A^-1 = [1/det(A)] adj(A) =

-5

-2

5

2

1

-2

-4

-3

5

5. (a) The coefficient matrix of the given linear system is A =

1

3

2

2

7

7

2

5

2

Let B =

1

3

2

1

0

0

2

7

7

0

1

0

2

5

2

0

0

1

We will reduce B to its RREF as under:

Add -2 times the 1st row to the 2nd row

Add -2 times the 1st row to the 3rd row

Add 1 times the 2nd row to the 3rd row

Add -3 times the 3rd row to the 2nd row

Add -2 times the 3rd row to the 1st row

Add -3 times the 2nd row to the 1st row

Then the RREF of B is

1

0

0

-21

4

7

0

1

0

10

-2

-3

0

0

1

-4

1

1

Hence A^-1=

-21

4

7

10

-2

-3

-4

1

1

Let X = (x₁,x₂,x₃)^T. Then X = A^-1b = A^-1(2,-1,7)^T= (3,1,-2)^T. Hence, x₁ = 3, x₂ = 1 and x₃ = -2.

(b) We will use Cramer’s rule to compute x₁,x₂ and x₃.

D = det(A) = 1, D_x1 = 3,D_x2 = 1 and D_x3 = -2. Then x₁ = D_x1/D = 3, x₂ = D_x2/D =1 and x₃ = D_x3/D = -2.

1	5	1	1	0	0
2	5	0	0	1	0
2	7	1	0	0	1