Adjoint of Matrix Questions 3
In this page adjoint of matrix questions 3 we are going to see solution of question 1 based on the topic ad-joint of matrix.
Question 3
Find the ad-joint of the following matrix
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Solution:
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minor of 6 |
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= [4-3] = (1) = 1 | ||||||||||
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Cofactor of 6 |
= + (1) = 1 |
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minor of 2 |
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= [12-10] = (2) = 2 | ||||||||||
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Cofactor of 2 |
= - (2) = -2 |
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minor of 3 |
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= [9-10] = (-1) = -1 | ||||||||||
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Cofactor of 3 |
= + (-1) = -1 |
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minor of 3 |
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= [8-9] = (-1) = -1 | ||||||||||
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Cofactor of 3 |
= - (-1) = 1 |
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minor of 1 |
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= [24-30] = (-6) = -6 | ||||||||||
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Cofactor of 1 |
= + (-6) = -6 |
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minor of 1 |
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= [18-20] = (-2) = -2 | ||||||||||
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Cofactor of 1 |
= - (-2) = 2 |
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minor of 10 |
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= [2-3] = (-1) = -1 | ||||||||||
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Cofactor of 10 |
= + (-1) = -1 |
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minor of 3 |
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= [6-9] = (-3) = -3 | ||||||||||
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Cofactor of 3 |
= - (-3) = 3 |
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minor of 4 |
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adjoint of matrix questions 3 |
= [6-6] = (0) = 0 | |||||||||
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Cofactor of 4 |
= + (0) = 0 adjoint of matrix questions 3 |
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co-factor matrix = |
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adjoint of matrix= |
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- Types of matrices
- Equality of matrices
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- Algebraic properties of matrices
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- Cramer's Rule for 3 equations
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- Rank Method in Matrix
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- Linear dependence of vectors
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- Characteristic Equation of matrix
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