Help required in finding determinant using characteristic equation.

Question

A problem in my textbook proceeds as follows:

Find $\det(A)$ given that $A$ has $p(\lambda)$ as its characteristic polynomial

a) $p(\lambda)=\lambda^3-2\lambda^2+\lambda+5$

b) $p(\lambda)=\lambda^4-\lambda^3+7$

What I did was:

a) Since $\det(\lambda I-A)=\lambda^3-2\lambda^2+\lambda+5$, then $\det(-A)=5$. Hence, $\det(A)=-5$. b) Since $\det(\lambda I-A)=\lambda^4-2\lambda^3+7$, then $\det(-A)=7$. Hence, $\det(A)=7$.

Am I correct here?

Answer 1

You are almost correct but not quite.

Hint. If $$A=\pmatrix{1&2\cr 1&1\cr}$$ write down $-A$ and find the determinants of $A$ and $-A$, doing all the working and not attempting to use any short cuts. Then repeat for $$A=\pmatrix{1&2&3\cr 1&1&2\cr -1&0&1\cr}\ .$$ What do you notice?

Answer 2

Let $A = I$.

Then $\det(\lambda I - A) = \det [(\lambda - 1) I] = (\lambda - 1)^n = \ldots + (-1)^n$, for $A$ being $n\times n$.

By your logic, we'd conclude that $\det (-A) = (-1)^n$. But it's nonsense to say that. $A$ is the identity; the determinant of $I$ does not depend on its dimensionality.

Moreover, does it make sense to say that $\det(-I) = \det I$? I don't think so.

Answer 3

In general when the characteristic polynomial is written in simplest form the constant term of the polynomial is equal to ${-1}^n \cdot det(A)$. Where $n$ is the dimension of the matrix.