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        what is proof of
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

SAGAR SINGH - IIT DELHI
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 i ≠ j 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.

Sagar Singh
B.Tech IIT Delhi


7 years ago
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