Question

question 6
for which values of ( a ) is ( vec{x} = \begin{bmatrix} 1 \\ a end{bmatrix} ) an eigenvector of ( \begin{bmatrix} 1 & 0 \\ 2 & 2 end{bmatrix} )?
(a) ( vec{x} ) is an eigenvector only when ( a = -\frac{1}{2} ).
(b) ( vec{x} ) is an eigenvector only when ( a = 0 ).
(c) ( vec{x} ) is an eigenvector only when ( a = -2 ).
(d) ( vec{x} ) is an eigenvector for any ( a in mathbb{r} ).
(e) none of the above.

Explanation:

Step1: Recall Eigenvector Definition

An eigenvector \(\vec{X}\) of a matrix \(A\) satisfies \(A\vec{X}=\lambda\vec{X}\) for some scalar \(\lambda\). Let \(A =

$$\begin{bmatrix}1&0\\2&2\end{bmatrix}$$

\) and \(\vec{X}=

$$\begin{bmatrix}1\\a\end{bmatrix}$$

\). First, compute \(A\vec{X}\).

Step2: Compute \(A\vec{X}\)

Using matrix multiplication:

$$ A\vec{X}=
LATEXBLOCK2

LATEXBLOCK3
=
LATEXBLOCK4
=
LATEXBLOCK5
$$

Step3: Compute \(\lambda\vec{X}\)

Let \(\lambda\) be the eigenvalue. Then \(\lambda\vec{X}=\lambda

$$\begin{bmatrix}1\\a\end{bmatrix}$$

=

$$\begin{bmatrix}\lambda\\\lambda a\end{bmatrix}$$

\).

Step4: Equate Components

Set \(A\vec{X}=\lambda\vec{X}\). So we have the system:

1. \(1=\lambda\) (from the first component)
2. \(2 + 2a=\lambda a\) (from the second component)

Substitute \(\lambda = 1\) into the second equation: \(2 + 2a=1\cdot a\).

Step5: Solve for \(a\)

Simplify \(2 + 2a=a\):
Subtract \(a\) from both sides: \(2 + a = 0\)
Subtract \(2\) from both sides: \(a=-2\). Wait, but let's check again. Wait, no—wait, maybe I made a mistake. Wait, no, let's re - do the matrix multiplication. Wait, \(A\vec{X}=

$$\begin{bmatrix}1*1 + 0*a\\2*1+2*a\end{bmatrix}$$

=

$$\begin{bmatrix}1\\2 + 2a\end{bmatrix}$$

\). And \(\lambda\vec{X}=\lambda

$$\begin{bmatrix}1\\a\end{bmatrix}$$

=

$$\begin{bmatrix}\lambda\\\lambda a\end{bmatrix}$$

\). From the first component, \(\lambda = 1\). Then from the second component: \(2 + 2a=\lambda a=1*a\). So \(2+2a=a\) implies \(2=a - 2a=-a\), so \(a=-2\). But wait, the options: option C is \(a = - 2\). But wait, let's check again. Wait, maybe I messed up the matrix. Wait, the matrix is \(

$$\begin{bmatrix}1&0\\2&2\end{bmatrix}$$

\), vector \(

$$\begin{bmatrix}1\\a\end{bmatrix}$$

\). So \(A\vec{X}=

$$\begin{bmatrix}1\\2 + 2a\end{bmatrix}$$

\), \(\lambda\vec{X}=

$$\begin{bmatrix}\lambda\\\lambda a\end{bmatrix}$$

\). So first component: \(1=\lambda\), second component: \(2 + 2a=\lambda a\). So with \(\lambda = 1\), \(2+2a=a\) gives \(a=-2\). So option C says \(a=-2\). But wait, let's check if this is correct. Let's plug \(a = - 2\) into \(\vec{X}=

$$\begin{bmatrix}1\\-2\end{bmatrix}$$

\), and compute \(A\vec{X}\): \(

$$\begin{bmatrix}1&0\\2&2\end{bmatrix}$$

$$\begin{bmatrix}1\\-2\end{bmatrix}$$

=

$$\begin{bmatrix}1\\2-4\end{bmatrix}$$

=

$$\begin{bmatrix}1\\-2\end{bmatrix}$$

\), which is \(1\cdot\vec{X}\), so \(\lambda = 1\). So it works. Wait, but earlier when I thought I made a mistake, but no, this is correct. So the value of \(a\) is \(-2\), which is option C.

Wait, but wait, let's check the options again. Option C: \(\vec{X}\) is an eigenvector only when \(a=-2\). So the answer should be C.

Answer:

C. \(\vec{X}\) is an eigenvector only when \(a = - 2\).