# 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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