If `A = [(1,2), (-1, 4)]` and `theta` be the angle between the two non-zero column vectors `X` such that `AX = lambda X` for some scalar `lambda,` then `9 sec^2 theta` is equal to

Options

A. 13
B. 12
C. 11
D. 10

Answer

To solve the problem step by step, we will find the eigenvalues of the matrix \( A \) and then use them to find the angle \( \theta \) between the corresponding eigenvectors. Here's the detailed solution:

### Step 1: Define the matrix \( A \)
The given matrix is:
\[ A = \begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix} \]

### Step 2: Set up the characteristic equation
To find the eigenvalues, we need to solve the characteristic equation:
\[ \text{det}(A - \lambda I) = 0 \]
where \( I \) is the identity matrix. Thus,
\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 2 \\ -1 & 4 - \lambda \end{pmatrix} \]

### Step 3: Compute the determinant
The determinant is calculated as follows:
\[ \text{det}(A - \lambda I) = (1 - \lambda)(4 - \lambda) - (-1)(2) \]
Expanding this gives:
\[ = (1 - \lambda)(4 - \lambda) + 2 = 4 - 5\lambda + \lambda^2 + 2 = \lambda^2 - 5\lambda + 6 \]

### Step 4: Solve the characteristic equation
Setting the determinant to zero:
\[ \lambda^2 - 5\lambda + 6 = 0 \]
Factoring the quadratic:
\[ (\lambda - 2)(\lambda - 3) = 0 \]
Thus, the eigenvalues are:
\[ \lambda_1 = 2, \quad \lambda_2 = 3 \]

### Step 5: Find the eigenvectors
#### For \( \lambda_1 = 2 \):
Substituting \( \lambda = 2 \) into \( A - \lambda I \):
\[ A - 2I = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \]
Setting up the equation:
\[ \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \]
This gives us:
\[ -v_1 + 2v_2 = 0 \implies v_1 = 2v_2 \]
Choosing \( v_2 = 1 \), we get the eigenvector:
\[ \mathbf{v_1} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \]

#### For \( \lambda_2 = 3 \):
Substituting \( \lambda = 3 \):
\[ A - 3I = \begin{pmatrix} -2 & 2 \\ -1 & 1 \end{pmatrix} \]
Setting up the equation:
\[ \begin{pmatrix} -2 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \]
This gives us:
\[ -2v_1 + 2v_2 = 0 \implies v_1 = v_2 \]
Choosing \( v_2 = 1 \), we get the eigenvector:
\[ \mathbf{v_2} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]

### Step 6: Find the angle \( \theta \)
The cosine of the angle \( \theta \) between the two vectors can be found using the formula:
\[ \cos \theta = \frac{\mathbf{v_1} \cdot \mathbf{v_2}}{|\mathbf{v_1}| |\mathbf{v_2}|} \]

#### Calculate the dot product:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = 2 \cdot 1 + 1 \cdot 1 = 2 + 1 = 3 \]

#### Calculate the magnitudes:
\[ |\mathbf{v_1}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]
\[ |\mathbf{v_2}| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

#### Substitute into the cosine formula:
\[ \cos \theta = \frac{3}{\sqrt{5} \cdot \sqrt{2}} = \frac{3}{\sqrt{10}} \]

### Step 7: Find \( \sec^2 \theta \)
Since \( \sec \theta = \frac{1}{\cos \theta} \):
\[ \sec \theta = \frac{\sqrt{10}}{3} \]
Thus,
\[ \sec^2 \theta = \left(\frac{\sqrt{10}}{3}\right)^2 = \frac{10}{9} \]

### Step 8: Calculate \( 9 \sec^2 \theta \)
Finally,
\[ 9 \sec^2 \theta = 9 \cdot \frac{10}{9} = 10 \]

### Final Answer:
\[ \boxed{10} \]