# Find All Values of $a$ which Will Guarantee that $A$ Has Eigenvalues 0, 3, and -3.

## Problem 719

Let $A$ be the matrix given by
$A= \begin{bmatrix} -2 & 0 & 1 \\ -5 & 3 & a \\ 4 & -2 & -1 \end{bmatrix}$ for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.

## Solution.

Let $p(t)$ be the characteristic polynomial of $A$, i.e. let $p(t)=\det(A-tI)=0$. By expanding along the second column of $A-tI$, we can obtain the equation
\begin{align*}
p(t)
&=
\det\left(
\begin{bmatrix}
-2 & 0 & 1 \\
-5 & 3 & a \\
4 & -2 & -1
\end{bmatrix}

\begin{bmatrix}
t & 0 & 0 \\
0 & t & 0 \\
0 & 0 & t
\end{bmatrix}
\right)
\\
&=
\begin{vmatrix}
-2-t & 0 & 1 \\
-5 & 3-t & a \\
4 & -2 & -1-t
\end{vmatrix}
\\
&=
(3-t)
\begin{vmatrix}
-2-t & 1 \\
4 & -1-t
\end{vmatrix}
+2
\begin{vmatrix}
-2-t & 1 \\
-5 & a
\end{vmatrix}
\\
&=
(3-t)
\left[(-2-t)(-1-t)-4\right] +2\left[(-2-t)a+5\right] \\
&=
(3-t)
(2+t+2t+t^{2}-4)
+2(-2a-t a+5)
\\
&=
(3-t)(t^{2}+3t-2)
+(-4a-2t a+10)
\\
&=
3t^{2}+9t-6-t^{3}-3t^{2}+2t
-4a-2t a+10
\\
&=
-t^{3}+11t-2t a+4-4a
\\
&=
-t^{3}+(11-2a)t+4-4a
.
\end{align*}
For the eigenvalues of $A$ to be $0$, $3$, and $-3$, the characteristic polynomial $p(t)$ must have roots at $t=0,3,-3$. This implies
\begin{align*}
p(t)
=
– t(t-3)(t+3)
=
– t(t^{2}-9)
=
– t^{3}+ 9t.
\end{align*}

Therefore,
$-t^{3}+(11-2a)t+4-4a =-t^{3}+9t .$

For this equation to hold, the constant terms on the left and right hand sides of the above equation must be equal. This means that $4-4a=0$, which implies $a=1$. Hence $A$ has eigenvalues $0,3,-3$ precisely when $a=1$.

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