inversion method questions 2

Inversion Method Questions 2

In this page inversion method questions 2 we are going to see how to solve the given linear equations using this particular method in matrices.

Question 2:

Solve the following linear equation by inversion method

x + 2y + z = 7

2x - y + 2z = 4

x + y - 2z = -1

Solution:

First we have to write the given equation in the form AX = B. Here X represents the unknown variables. A represent coefficient of the variables and B represents constants.

1	2	1
2	-1	2
1	1	-2

-1

To solve this we have to apply the formula X = A⁻¹ B

|A|

1	2	1
2	-1	2
1	1	-2

|A|

= 2

-1		2

1		-2

-2

2		2

1		-2

2		-1

1		1

|A| = 1 [2-2] - 2 [-4-2] +1 [2+1]

= 1 [0] - 2 [-6] +1 [3]

= 0 + 12 + 3

= 15

|A| = 15 ≠ 0

Since A is a non singular matrix. A⁻¹ exists. inversion method questions 2

minor of 1

-1	2
1	-2

inversion method questions 2

= [2-2]

= 0

Cofactor of 1

= + (0)

= 0

minor of 2

2	2
1	-2

= [-4-2]

= -6

Cofactor of 2

= - (-6)

= 6

minor of 1

2	-1
1	1

= [2-(-1)]

= [2+1]

= 3

Cofactor of 1

= + (3)

= 3

minor of 2

2	1
1	-2

= [-4-1]

= -5

Cofactor of 2

= - (-5)

= 5

minor of -1

1	1
1	-2

= [-2-1]

= -3

Cofactor of -1

= + (-3)

= -3

minor of 2

1	2
1	1

= [1-2]

= -1

Cofactor of 2

= - (-1)

= 1

minor of 1

2	1
-1	2

= [4-(-1)]

= [4+1]

= 5

Cofactor of 1

= + (5)

= 5

minor of 1

1	1
2	2

= [2-2]

= [0]

= 0

Cofactor of 1

= - (0)

= 0

minor of -2

1	2
2	-1

inversion method questions 2

= [-1-4]

= [-5]

= -5

Cofactor of -2

= + (-5)

= -5

co-factor matrix =

0	6	3
5	-3	1

5	0	-5

adjoint of matrix=

0	5	5
6	-3	0

3	1	-5

A⁻¹ = 1/15

0	5	5
6	-3	0

3	1	-5

= 1/15

0	5	5
6	-3	0

3	1	-5

-1

= 1/15

-3

-1

-5

-1

=1/15

(0+20-5)

(42-12-0)

(21+4+5)

=1/15

(15)

(30)

Solution:

x = 1

y = 2

z = 2

Inversion Method Question2 to Inversion Method

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