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vikash chandola Grade:

what is proof of

A*adj(A) = |A|*I

where A is a square matrix of order n and I is identity matrix of order n. plz reply with proper derivation.


7 years ago

Answers : (1)

879 Points

Dear vikash,

As a consequence of Laplace formula for the determinant of an n×n matrix A, we have

\mathbf{A}\, \mathrm{adj}(\mathbf{A}) = \mathrm{adj}(\mathbf{A})\, \mathbf{A} = \det(\mathbf{A})\, \mathbf{I}\qquad (*)

where I is the n×n identity matrix Indeed, the (i,i) entry of the product A adj(A) is the scalar product of row i of A with row i of the cofactor matrix C, which is simply the Laplace formula for det(A) expanded by row i. Moreover, for ij the (i,j) entry of the product is the scalar product of row i of A with row j of C, which is the Laplace formula for the determinant of a matrix whose i and j rows are equal and is therefore zero.

From this formula follows one of the most important results in matrix algebra: A matrix A over a commutative ring R is invertible if and only if det(A) is invertible in R.

For if A is an invertible matrix then

1 = \det(\mathbf I) = \det(\mathbf{A} \mathbf{A}^{-1}) = \det(\mathbf{A}) \det(\mathbf{A}^{-1}),

and if det(A) is a unit then (*) above shows that

\mathbf{A}^{-1} = \det(\mathbf{A})^{-1}\, \mathrm{adj}(\mathbf{A}).

We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech IIT Delhi

7 years ago
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