# Determinant of a matrix $P$ such that $AP=BP$.

Let $A,B$ be the two $3 \times 3$ matrices

$$A=\begin{bmatrix} 3 & 4 &0\\0 &-1&0\\4 &4 & -1 \end{bmatrix}$$

$$B= \begin{bmatrix} -1 & 1 &0\\0 &-1&0\\0 &0 & 3 \end{bmatrix}$$

Suppose that there is matrix $P$ that $AP=BP$. Prove that $|P|=0$.

My solution is very simple and I'm not sure if its currect.

$$|A|=-3 , |B|=3$$ $$-3|P|=3|P|$$ $$|P|=0$$

Other sultions are much longer. Is there a reason my prove is not correct?

• I haven't confirmed the determinants of $A$ and $B$, but if they are $3$ and $-3$, then you're right. Sep 3, 2016 at 12:24

Your proof would be correct$^{[1]}$ if your determinants were right ! But for me, $\det A =3 = \det B$. See (A) and (B).
You can actually see that if $\det P \neq 0$, then $P$ would be invertible, which would yield $APP^{-1}=A=B=BPP^{-1}$, a contradiction.
$^{[1]}$ provided that your matrices have real coefficients, and not in $\Bbb Z/6\Bbb Z$.