Find $\det(A^3B^2+A^2B^3)$ for $A,B$ s.t. $AB = BA$ given $\det A,\det B$.

Question

If two matrices $A$ and $B$ are such that $AB = BA$ with $\det A = 1$ and $\det B = 0$, then what is $\det(A^3B^2 + A^2B^3)$ ?

Thanks man...!! Can you help me with this question ? – Soham Sarkar Jul 22 '17 at 06:51 — Soham Sarkar, Jul 22 '17 at 06:51

score 5 · Answer 1 · answered Jul 22 '17 at 06:49

5

We have that: $$\det(A^3B^2 + A^2B^3) = \det(A^2B^2(A + B)) = \det(A^2B^2)\det(A + B) = (1)^2(0)^2\det(A + B) = 0$$

answered Jul 22 '17 at 06:49

Mark Schultz-Wu

6,986

score 5 · Accepted Answer · answered Jul 22 '17 at 06:57

5

Even without assuming commutativity, we see that $$ A^3B^2+A^2B^3=(A^3B+A^2B^2)\cdot B$$ is singular.

answered Jul 22 '17 at 06:57

Hagen von Eitzen

374,180

How do we understand that the above relation of matrices is singular ? – Soham Sarkar Jul 22 '17 at 07:05
2

@Soham: The kernel of $B $ is nontrivial (since $\det B=0$), and thus so is the kernel of $CB $, for any matrix $C $. – PhoemueX Jul 22 '17 at 07:16

score 2 · Answer 3 · answered Jul 22 '17 at 06:50

2

\begin{align*} \text{det}(A^3B^2+A^2B^3) & = \text{det}A^2 \text{det} B^2\text{det}(A+B)\\ & = (\text{det}A)^2 (\text{det} B)^2\text{det}(A+B)\\ &=0. \end{align*}

answered Jul 22 '17 at 06:50

Anurag A

41,067

Could we get the value of the det if det B would not be 0 ? – Soham Sarkar Jul 22 '17 at 07:02
@SohamSarkar probably not without some additional information. – Anurag A Jul 22 '17 at 17:21

Find $\det(A^3B^2+A^2B^3)$ for $A,B$ s.t. $AB = BA$ given $\det A,\det B$.

3 Answers3