Question

The determinant of a \(2 \times 2\) square matrix is 1.6. If the sum of the eigenvalues is \(-2.8\), the eigenvalues are

• A. \(-2\) and \(-0.8\)
• B. \(-1.8\) and \(-1\)
• C. \(-3\) and \(0.2\)
• D. \(-1.2\) and \(-1.6\)

Hint:

Set up the characteristic equation from the trace and determinant of the matrix.

Answer 1

The Correct Option is A

Step 1: Write down what is given about the eigenvalues.
For a \(2\times2\) matrix, the sum of the eigenvalues equals the trace, and the product of the eigenvalues equals the determinant.
So \(\lambda_1 + \lambda_2 = -2.8\) and \(\lambda_1 \lambda_2 = 1.6\).

Step 2: Form the characteristic quadratic.
The eigenvalues satisfy \(\lambda^2 - (\lambda_1+\lambda_2)\lambda + \lambda_1\lambda_2 = 0\), which gives \(\lambda^2 + 2.8\lambda + 1.6 = 0\).

Step 3: Solve the quadratic.
Discriminant \(= 2.8^2 - 4(1.6) = 7.84 - 6.4 = 1.44\), and \(\sqrt{1.44} = 1.2\).
\(\lambda = \dfrac{-2.8 \pm 1.2}{2}\), giving \(\lambda = -0.8\) or \(\lambda = -2\).

Step 4: Check the wrong options.
Options B, C, D each fail either the sum or the product test; for example, \(-1.8 \times -1 = 1.8 \ne 1.6\).

Final Answer:
The eigenvalues are \(-2\) and \(-0.8\), so option A is correct. \[ \boxed{\lambda = -2,\ -0.8} \]