If A and B are two matrices then determinant(A*B)=determinant(A)*determinant(B).Is the above tue.If yes please prove it.

Dear sudarshan gupta,

the above statement is true if A, B are square matrices of same order.

the proof is below.

We would like to prove that det(AB) = det(A)×det(B), for any two matrices A and B.

To start, assume B is a singular matrix, with determinant 0.  In other words, the vectors of B do not span the entire space.  If A is another matrix, each row in A*B is a linear combination of the rows of B.  The space spanned by A*B is also spanned by B.  This is a proper subspace, not the entire space, hence A*B is singular and its determinant is 0.

Next assume B is nonsingular, hence it has a nonzero determinant.  Premultiply B by a series of elementary matrices that perform the elementary operations associated with gaussian elimination.  Thus t = e1e2e3e4...ejB, where t is the upper triangular matrix that results from gaussian elimination.  Since the determinant is nonzero, the main diagonal of t contains all nonzero entries.

Premultiply t by yet more elementary matrices to perform back substitution.  In other words, scaled versions of the bottom row are subtracted from all the others, clearing out the last column.  Then the second to last column is cleared, and so on, until a diagonal matrix is left.  Write this as d = e1e2e3...ejB, where d is the diagonal matrix.  The determinant of d is the determinant of B times the determinants of the elementary matrices.  Of course most of these determinants are 1, except for the matrices that swap two rows.

Verify that every elementary matrix has an inverse, and let fi be the inverse of ei.  These inverses are also elementary matrices, hence det(fi*M) = det(fi)×det(M).  Now B = f1f2f3...fjd.  Replace A*B with A times a bunch of elementary matrices times d, and fold each elementary matrix into A, one at a time.  Each time the determinant commutes with multiplication.  This also holds for d.  When everything is multiplied together, det(A*B) = det(A)×det(B).

For any two matrices A and B, the determinant of the product is the product of the determinants.  Restrict attention to nonsingular matrices, and the determinant implements a group homomorphism into the base ring.

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