Let A and B be n ×n matrices. Is det (AB ) = det (BA ) ? Explain your reasoning.?

Question

Let A and B be n ×n matrices. Is det (AB ) = det (BA ) ? Explain your reasoning.?

Answer 1

You can answer this question!!!

Take on matrix with R1 = a b, R2 = c d and secod with R1 = p q and R2 with r s and try out. You will find det of one product contains a term (ap+br)(cq+ds) while det of other product contains a term

(ap+cq)( rb+sd) which are unequal.

Check it please!

Answer 2

Yes.

Case 1: Suppose A is invertible. Then we have

det(B) = det(B)

Multiply both sides by det(AA^-1):

det(AA^-1) det(B) = det(AA^-1) det(B)

Apply the fact that determinants work nicely with multiplication:

det(A) det(A^-1) det(B) = det(A) det(A^-1) det(B)

Since these determinants are all just scalars, we can use the commutativity of multiplication:

det(A^-1) det(A) det(B) = det(A^-1) det(B) det(A)

Now again apply the fact that determinants work nicely with multiplication:

det(A^-1) det (AB) = det(A^-1) det (BA)

Now divide by det(A^-1), which is nonzero because A^-1 is invertible. We get:

det (AB) = det (BA).

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Case 2: Suppose that A is not invertible.

Then consider the matrix A + Ix, where I is the identity matrix.

Notice that det (A + Ix) is a nonzero polynomial in x of degree n, and therefore it has at most n roots. If we let x be a number that's not a root of det (A + Ix), then we can carry out reasoning similar to the above:

det(B) = det (B)

det ((A + Ix)^-1 (A + Ix)) det(B) = det ((A + Ix)^-1 (A + Ix)) det (B)

det ((A + Ix)^-1) det(A + Ix) det (B) = det ((A + Ix)^-1) det (A + Ix) det (B)

det ((A + Ix)^-1) det(A + Ix) det (B) = det ((A + Ix)^-1) det (B) det (A + Ix)

det ((A + Ix)^-1) det((A + Ix)B) = det((A + Ix)^-1) det (B(A + Ix))

det ((A + Ix)B) = det(B(A + Ix)).

Since this last equation is an equation of polynomials of degree at most n, and it holds for all but finitely many x, then it is forced to hold for all x as well. (Two polynomials of degree n or less that agree at n + 1 positions or more have to be equal.)

So, in particular, the polynomials are equal when x = 0, which gives

det (AB) = det (BA).