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.
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To solve this we have to apply the formula X = A⁻¹ B
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|A| |
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|A| |
= 2 |
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-2 |
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+1 |
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|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
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minor of 1 |
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inversion method questions 2 |
= [2-2] = 0 | |||||||||
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Cofactor of 1 |
= + (0) = 0 | |||||||||
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minor of 2 |
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= [-4-2] = -6 | ||||||||||
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Cofactor of 2 |
= - (-6) = 6 | |||||||||
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minor of 1 |
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= [2-(-1)] = [2+1] = 3 | ||||||||||
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Cofactor of 1 |
= + (3) = 3 | |||||||||
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minor of 2 |
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= [-4-1] = -5 | ||||||||||
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Cofactor of 2 |
= - (-5) = 5 | |||||||||
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minor of -1 |
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= [-2-1] = -3 | ||||||||||
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Cofactor of -1 |
= + (-3) = -3 | |||||||||
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minor of 2 |
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= [1-2] = -1 | ||||||||||
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Cofactor of 2 |
= - (-1) = 1 | |||||||||
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minor of 1 |
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= [4-(-1)] = [4+1] = 5 | ||||||||||
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Cofactor of 1 |
= + (5) = 5 | |||||||||
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minor of 1 |
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= [2-2] = [0] = 0 | ||||||||||
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Cofactor of 1 |
= - (0) = 0 | |||||||||
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minor of -2 |
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inversion method questions 2 |
= [-1-4] = [-5] = -5 | |||||||||
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Cofactor of -2 |
= + (-5) = -5 |
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co-factor matrix = |
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adjoint of matrix= |
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A⁻¹ = 1/15 |
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= 1/15 |
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= 1/15 |
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Solution:
x = 1
y = 2
z = 2
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