Determinant

Algebra Level 3

If $A$ and $B$ are different non-singular square matrices, satisfying $A^3=B^3$ and $A^2B=B^2A$ , then choose the correct option.

Nothing can be concluded

\det(A^2+B^2)=0

but

\det(A-B) \neq 0

\det(A^2+B^2)=\det(A-B)=0

\det(A-B)=0

but

\det(A^2+B^2) \neq 0

both

\det(A^2+B^2)

and

\det(A-B) \neq 0

1 solution

Mark Hennings
Feb 4, 2020

Since $A^2B = B^2A$ and $A^3=B^3$ we see that $0 = A^3 - B^3 = (A^2+B^2)(A-B)$ .

If $A^2 + B^2$ is non-singular then (multiplying by its inverse) $A-B=0$ , and so $A=B$ , which is not possible. Thus $|A^2+B^2|=0$ .
If $A-B$ is non-singular then, similarly, $A^2+B^2=0$ , so that $B^2 = -A^2$ , and hence $A^2B = B^2A = -A^3$ , so that ( $A$ is non-singular) $B=-A$ , and hence $A^3 = B^3 = -A^3$ , so that $A^3=0$ , and hence $A=0$ , which is not possible. Thus $|A-B|=0$ as well.

Where you say "non-null" you need to say "non-singular". The only "null" matrix is the matrix with only zeros as entries. Being non-null isn't sufficient to be invertible. "Non-singular" is the adjective you want.

Richard Desper - 1 year, 4 months ago

I agree. In my discussions with the author about this question, I worked with his usage of non-null, concentrating on the other issues with the question, and did not correct this one.. I have corrected my solution.

Mark Hennings - 1 year, 4 months ago

Determinant

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