Question

The eigenvalues of the matrix \( P = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix} \) are

• A. \(-1\) and \(-2\)
• B. \(1\) and \(2\)
• C. \(0\) and \(-1\)
• D. \(0\) and \(1\)

Hint:

For a \(2\times2\) matrix, the eigenvalues are the roots of \(\lambda^2 - (\text{trace})\lambda + \det = 0\).

Answer 1

The Correct Option is A

Step 1: Write down the characteristic equation.
The eigenvalues \(\lambda\) of a matrix \(P\) satisfy \(\det(P - \lambda I) = 0\).
For \( P = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix} \), this becomes
\[ \det \begin{pmatrix} 1-\lambda & -2 \\ 3 & -4-\lambda \end{pmatrix} = 0 \]

Step 2: Expand the determinant.
\[ (1-\lambda)(-4-\lambda) - (-2)(3) = 0 \]
\[ (1-\lambda)(-4-\lambda) + 6 = 0 \]
Expand the product: \( (1-\lambda)(-4-\lambda) = -4 -\lambda +4\lambda +\lambda^2 = \lambda^2 +3\lambda -4 \).
So the equation becomes \[ \lambda^2 + 3\lambda - 4 + 6 = 0 \] \[ \lambda^2 + 3\lambda + 2 = 0 \]

Step 3: Use the trace-determinant shortcut to check.
For a \(2\times 2\) matrix, the characteristic equation is \(\lambda^2 - (\text{trace})\lambda + \det(P) = 0\).
Trace of \(P\) is \(1 + (-4) = -3\). Determinant of \(P\) is \(1(-4) - (-2)(3) = -4+6 = 2\).
So the equation is \(\lambda^2 - (-3)\lambda + 2 = 0\), which is \(\lambda^2 + 3\lambda + 2=0\). This matches Step 2, so the expansion is confirmed to be correct.

Step 4: Factor and solve.
\[ \lambda^2 + 3\lambda + 2 = (\lambda+1)(\lambda+2) = 0 \]
So \(\lambda = -1\) or \(\lambda = -2\).

Step 5: Check why the other options are wrong.
Option (B) 1 and 2 would need the trace to be \(3\), but the actual trace is \(-3\), so this option is wrong.
Options (C) and (D) include a zero eigenvalue, which can only happen when \(\det(P)=0\); here \(\det(P)=2 \neq 0\), so neither option is correct.

Final Answer:
The eigenvalues of \(P\) are \(-1\) and \(-2\).
\[ \boxed{-1 \text{ and } -2} \]