Trace and determinant of characteristic polynomial.

Question

I Just started learning linear algebra. In my homework exercise i have this question: The characteristic polynomial of a square matrix $A$ of order $3$ is $|\lambda I-A|=\lambda^{3}+3\lambda^{2}+4\lambda-3$ Let $x=$ Trace$(A)$ and $y=|A|$, Then ,

(A) $\dfrac{x}{y}=\dfrac{3}{4}$

(B) $\dfrac{x}{y}=\dfrac{4}{3}$

(C) $x=y=-3$

(D) $x=3$ and $y=-3$

I usually find sum and product of roots using characteristic polynomial. . My trace when i calculated is $-3$ (sum of roots) and $|A|=3$(product of roots) But it doesnt matches with options. Can someone tell me where did i do wrong ?

Answer 1

It is just a misconception of sign i.e. $(-1)^3\{\lambda^{3}-(Trace(A))\lambda^{2}+\lambda-|A|\}$ = $|\lambda I - A| $. So your last option is correct.