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The characteristic polynomial of a $3\times3$ matrix $A$ is given by

$$ \chi_A(x) = x^3 + ax^2 +bx +c $$

and takes the values $\chi_A(-1) = 4$, $\chi_A(2)=4$ and $\chi_A(-3) = -16$. Is $A$ invertible?

I got $0$ points at this question on a quiz and I want to understand why. What is the correct way to do this question? Thanks so much.

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4 Answers

up vote 3 down vote accepted


1) $\,\det A=\pm c\,$

2) $\,A\,$ is invertible iff $\,c\neq 0\,$

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By substituting the values you will get 3 equations with 3 variables namely $$ 4=-1+a-b+c$$ $$4=8+4a+2b+c$$ $$-16=-27+9a-3b+c$$ solving these equations you will get $a=0,b=-3,c=2$, Thus the charactirestic function is$$\chi_A=x^3-3x+2=(x-1)^2(x+2)$$ which implies the eigenvalues of $A$ are 1,1 and -2 which implies that

$Det(A)=1\times 1\times-2=-2\ne 0$ Which implies $A$ is invertible.

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The overall strategy is to use the following facts:

  1. A matrix is singular if and only if it has 0 as an eigenvalue;
  2. The eigenvalues of a matrix are the roots of its characteristic polynomial;
  3. The constant coefficient of a polynomial is the product of its roots.

So to start, we need to find the coefficients of $\chi$. Plugging your given data into the polynomial, we get a linear system in the coefficients:

$$\left[\begin{array}{ccc} 1 & -1 & 1\\4 & 2 & 1\\9 & -3 & 1\end{array}\right]\left[\begin{array}{c}a\\b\\c\end{array}\right] = \left[\begin{array}{c}5\\-4\\11\end{array}\right].$$

Solving this system gives $\chi(x) = x^3 -3x + 2$. Since the constant term is non-zero, all of the eigenvalues are non-zero, and so the matrix is invertible.

EDIT: Note that sometimes you can save some work by looking at the geometry of the data. For instance, if you had had $\chi(3) = -16$ instead, you would know that the three roots of the polynomial would have to be in the intervals $(-\infty, -1)$, $(2,3)$, and $(3,\infty)$, so $0$ could not possibly be a root. Unfortunately, for the data in this problem, it is plausible that $\chi$ has two roots on $(-1,2)$, including one at 0, so you have to do the full algebra.

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Perhaps if you provide us with your approach to the problem, we can help explain where you went wrong.

Without that here is a general idea. A square matrix is invertible if and only if it has trivial kernel. That is the only vector such that $Av = 0$ is $v = 0$. Think about the kernel as the set of all eigenvectors for the zero eigenspace.

So how you might think about this problem is that its asking you whether or not the 0 eigenspace (kernel) has any nonzero eigenvactors, equivalently its asking if $0$ is a root of the characteristic polynomial.

Given the data how can you find the actual characteristic polynomial? Figure that out, and you'll be able to tell if it has a nontrivial kernel or not.

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