# Quiz 11. Find Eigenvalues and Eigenvectors/ Properties of Determinants

## Problem 363

(a) Find all the eigenvalues and eigenvectors of the matrix
$A=\begin{bmatrix} 3 & -2\\ 6& -4 \end{bmatrix}.$

(b) Let
$A=\begin{bmatrix} 1 & 0 & 3 \\ 4 &5 &6 \\ 7 & 0 & 9 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}.$ Then find the value of
$\det(A^2B^{-1}A^{-2}B^2).$ (For part (b) without computation, you may assume that $A$ and $B$ are invertible matrices.)

## (a) Solution.

To determine eigenvalues of $A$, we compute the determinant of $A-\lambda I$.
We have
\begin{align*}
\det(A-\lambda I)&=\begin{vmatrix}
3-\lambda & -2\\
6& -4-\lambda
\end{vmatrix}\\
&=(3-\lambda)(-4-\lambda)+12\\
&=\lambda^2+\lambda=\lambda(\lambda+1).
\end{align*}
The eigenvalues are solutions of $\det(A-\lambda I)=0$, hence eigenvalues of $A$ are $0, -1$.

Next, we find the eigenvector corresponding to the eigenvalue $\lambda=0$.
Eigenvectors $\mathbf{x}$ are nonzero solutions of $(A-0I)\mathbf{x}=\mathbf{0}$.
Thus, we solve $A\mathbf{x}=\mathbf{0}$. The augmented matrix of the system is
\begin{align*}
\left[\begin{array}{rr|r}
3 & -2 & 0 \\
6 & -4 & 0
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rr|r}
3 & -2 & 0 \\
0 & 0 & 0
\end{array} \right] \xrightarrow{\frac{1}{3}R_1}
\left[\begin{array}{rr|r}
1 & -2/3 & 0 \\
0 & 0 & 0
\end{array} \right].
\end{align*}
Thus, if $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a solution, then $x_1=\frac{2}{3}x_2$, hence
$\mathbf{x}=x_2\begin{bmatrix} 2/3 \\ 1 \end{bmatrix}$ is an eigenvector corresponding to $\lambda=0$ for any nonzero scalar $x_2$.

Finally, we find the eigenvectors corresponding to the eigenvalue $\lambda=-1$.
In this case we need to solve $(A-(-1)I)\mathbf{x}=\mathbf{0}$.
The augmented matrix is
\begin{align*}
\left[\begin{array}{rr|r}
4 & -2 & 0 \\
6 & -3 & 0
\end{array} \right] \xrightarrow[\frac{1}{3}R_2]{\frac{1}{2}R_1} \left[\begin{array}{rr|r}
2 & -1 & 0 \\
2 & -1 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rr|r}
2 & -1 & 0 \\
0 & 0 & 0
\end{array} \right] \xrightarrow{\frac{1}{2}R_1}
\left[\begin{array}{rr|r}
1 & -1/2 & 0 \\
0 & 0 & 0
\end{array} \right].
\end{align*}
It follows that the eigenvectors associated to $\lambda=-1$ are
$\mathbf{x}=x_2\begin{bmatrix} 1/2 \\ 1 \end{bmatrix}$ for any nonzero scalar $x_2$.

## (b) Solution.

We use the following two properties of determinants.
Let $C$ and $D$ be $n\times n$ matrices. Then we have
\begin{align*}
\det(CD)&=\det(C)\det(D)
\end{align*}
and if $C$ is invertible, then
$\det(C^{-1})=\det(C)^{-1}=\frac{1}{\det(C)}.$

Using the properties of determinants, we compute
\begin{align*}
&\det(A^2B^{-1}A^{-2}B^2)\\&=\det(A)^2\det(B)^{-1}\det(A)^{-2}\det(B)^2\\
&=\det(A)^2\det(A)^{-2}\det(B)^{-1}\det(B)^2 && \text{(determinants are just numbers)}\\
&=\det(B).
\end{align*}

Hence it suffices to find the determinant of the matrix $B$.
Since the matrix $B$ is an upper triangular matrix, its determinant is the product of diagonal entries, thus
$\det(B)=2\cdot 3 \cdot 4=24.$ As a result, we obtain
$\det(A^2B^{-1}A^{-2}B^2)=24.$

## Comment.

These are Quiz 11 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

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