Question

Which one of the following matrices has eigenvalues 1 and 6?

• A. \( \begin{bmatrix} 5 & -2
  -2 & 2 \end{bmatrix} \)
• B. \( \begin{bmatrix} 3 & -1
  -2 & 2 \end{bmatrix} \)
• C. \( \begin{bmatrix} 3 & -1
  -1 & 2 \end{bmatrix} \)
• D. \( \begin{bmatrix} 2 & -1
  -1 & 3 \end{bmatrix} \)

Hint:

For \(2\times2\) matrices, always use trace and determinant to find eigenvalues quickly. Expanding the full determinant is often unnecessary.

Answer 1

The Correct Option is A

Step 1: Characteristic equation approach.
Eigenvalues \(\lambda\) satisfy: \[ \det(A - \lambda I) = 0 \] For eigenvalues 1 and 6, the polynomial should be \((\lambda-1)(\lambda-6) = \lambda^2 - 7\lambda + 6\). Step 2: Use trace and determinant shortcut.
For a \(2 \times 2\) matrix: \[ \text{Trace} = \lambda_1 + \lambda_2 = 1+6 = 7, \quad \det = \lambda_1 \lambda_2 = 6. \] Step 3: Check options.
- (A) Trace = \(5+2=7\). Det = \(5\cdot2 - (-2)(-2) = 10-4=6\). Matches perfectly
- (B) Trace = \(3+2=5\), Det = \(6-2=4\). Wrong.
- (C) Trace = \(3+2=5\), Det = \(6-1=5\). Wrong.
- (D) Trace = \(2+3=5\), Det = \(6-1=5\). Wrong. Step 4: Conclude.
Matrix (A) alone has trace 7 and determinant 6, hence eigenvalues 1 and 6. \[ \boxed{\begin{bmatrix} 5 & -2
-2 & 2 \end{bmatrix}} \]