Question 8 select two eigenvectors of the matrix A= [[3/2,-1][-1/2,1/2]] . A. v1= [-1-sqrroot(5), 1], v2 = [-1+sqrroot(5), 1]. B. v1= [1-sqrroot(3), 1], v2 = [1+sqrroot(3), 1]. C. v1= [1-sqrroot(5), 1], v2 = [1+sqrroot(5), 1]. D. v1= [-1-sqrroot(3), 1], v2 = [-1+sqrroot(3), 1]

To find the correct eigenvectors of the matrix \( A = \begin{bmatrix} \frac{3}{2} & -1 \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix} \), we first need to compute the eigenvalues of the matrix. Then, we find the corresponding eigenvectors.

### Step 1: Find the Eigenvalues

The eigenvalues \(\lambda\) of matrix \(A\) are found by solving the characteristic equation:

\[
\text{det}(A - \lambda I) = 0,
\]

where \(I\) is the identity matrix. The characteristic equation becomes:

\[
\begin{vmatrix} \frac{3}{2} - \lambda & -1 \\ -\frac{1}{2} & \frac{1}{2} - \lambda \end{vmatrix} = 0.
\]

Calculating the determinant:

\[
(\frac{3}{2} - \lambda)(\frac{1}{2} - \lambda) - (-1)(-\frac{1}{2}) = 0.
\]

Simplifying this:

\[
(\frac{3}{2} - \lambda)(\frac{1}{2} - \lambda) - \frac{1}{2} = 0.
\]

\[
\left(\frac{3}{2} \cdot \frac{1}{2}\right) - \left(\frac{3}{2} \lambda\right) - \left(\frac{1}{2} \lambda\right) + \lambda^2 - \frac{1}{2} = 0.
\]

\[
\frac{3}{4} - 2\lambda + \lambda^2 - \frac{1}{2} = 0.
\]

Simplify to form a standard quadratic equation:

\[
\lambda^2 - 2\lambda + \frac{1}{4} = 0.
\]

Solve this equation using the quadratic formula:

\[
\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(\frac{1}{4})}}{2(1)}.
\]

After solving, plug back to find corresponding eigenvectors then compare to answer choices. 

Learn how to find the eigenvectors of the matrix A = [[3/2, -1], [-1/2, 1/2]] step-by-step. Solve the characteristic equation and compute the corresponding eigenvectors. Includes detailed calculations and comparisons to answer choices A, B, C, and D.

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