One of the eigenvalues of A is where `A= [(2,5),(7,4)]`

Options

A. 1
B. 2
C. -3
D. 3

Answer

To find the eigenvalues of the matrix \( A = \begin{pmatrix} 2 & 5 \\ 7 & 4 \end{pmatrix} \), we will follow these steps:

### Step 1: Set up the characteristic equation
The eigenvalues of a matrix \( A \) can be found by solving the characteristic equation given by:

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

where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix.

### Step 2: Formulate the matrix \( A - \lambda I \)
For our matrix \( A \):

\[ A - \lambda I = \begin{pmatrix} 2 - \lambda & 5 \\ 7 & 4 - \lambda \end{pmatrix} \]

### Step 3: Calculate the determinant
Now, we need to calculate the determinant of the matrix \( A - \lambda I \):

\[ \text{det}(A - \lambda I) = (2 - \lambda)(4 - \lambda) - (5)(7) \]

### Step 4: Expand the determinant
Expanding the determinant:

\[ = (2 - \lambda)(4 - \lambda) - 35 \]
\[ = 8 - 2\lambda - 4\lambda + \lambda^2 - 35 \]
\[ = \lambda^2 - 6\lambda - 27 \]

### Step 5: Set the determinant to zero
Now, we set the determinant equal to zero to find the eigenvalues:

\[ \lambda^2 - 6\lambda - 27 = 0 \]

### Step 6: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -6 \), and \( c = -27 \):

\[ \lambda = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1} \]
\[ = \frac{6 \pm \sqrt{36 + 108}}{2} \]
\[ = \frac{6 \pm \sqrt{144}}{2} \]
\[ = \frac{6 \pm 12}{2} \]

### Step 7: Calculate the eigenvalues
This gives us two solutions:

1. \( \lambda_1 = \frac{18}{2} = 9 \)
2. \( \lambda_2 = \frac{-6}{2} = -3 \)

Thus, the eigenvalues of the matrix \( A \) are \( 9 \) and \( -3 \).

### Final Answer:
The eigenvalues of the matrix \( A \) are \( 9 \) and \( -3 \).

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