Let $M$ and $N$ be two 3x3 matrices of such that $MN=NM$. Further, $M\not =N^2$ and $M^2=N^4$, pick the righ-t options

Question

a)$|M^2+MN^2|=0$

b) There is a 3x3 non zero matrix $U$ such that $(M^2-MN^2)U$ is a zero matrix

c) There is a 3x3 matrix $U$ such that if $(M^2+MN^2)U$ is a zero matrix then $U$ is a zero matrix.

d) $|M^2+MN^2|\ge 1$

One solution I thought was $$M^2-N^4=O$$

$$(M-N^2)(M+N^2)=O$$

$$M+N^2=O$$ $$M^2+MN^2=O$$ Therefore their determinant is also zero. But I think it’s wrong because when $(M-N^2)(M+N^2)$ is expanded we get $M^2+MN^2-N^2M-N^4$. I know we have $MN=NM$, but that doesn’t satisfy the requirements.

As for the rest of the options, I don’t know how to work them out, at least not until I know this one.

Thanks

Answer 1

I can not believe no one gave a proper answer yet!

It is indeed true that: $$ MN^2 = N^2M$$

Proof:

$$ MN^2 = MNN = NMN= NNM = N^2 M$$

I have used the commutive property twice.

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b. Any matrix will do (proof: see equation you've derived)

c. contradicts b, hence false.

d. Contradicts a, hence false.

Answer 2

"As for the rest of the options, I don’t know how to work them out, at least not until I know this one. "

Why do you think so? You can always try some examples, say, diagonal matrices. For example, take $$ M=\operatorname{diag}(-1,1,1),\; N=\operatorname{diag}(1,-1,1) $$ Then $M^2=N^2=I$, so that $N^2=M^4$, but $M\neq N^2$. Obviously $$ M^2+MN^2=I+M=\operatorname{diag}(0,2,2) $$ has determinant zero. So $(d)$ cannot hold.

Concerning $(a)$: Take $M= \operatorname{diag}(-1,1,1)$ and $N= \operatorname{diag}(i,1,1)$. Then $M^2=I$, $N^4=I$ and $N^2=M$ sine $i^2=-1$. We have $$ M^2+MN^2=I+M^2=2I. $$ This has a non-zero determinant.