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: Recall the trace-determinant rule for a $2\times2$ matrix.
For any $2\times 2$ matrix, the sum of the two eigenvalues equals the trace, and the product of the two eigenvalues equals the determinant. This is much faster than solving the full characteristic equation when only the eigenvalues are needed.

Step 2: Compute the trace and determinant of $P$.
Given $P = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix}$:
trace $= 1 + (-4) = -3$.
determinant $= (1)(-4) - (-2)(3) = -4 + 6 = 2$.

Step 3: Check each option against these two conditions.
Option (A) $-1, -2$: sum $= -3$, which matches the trace, and product $= (-1)(-2) = 2$, which matches the determinant. Both conditions hold.
Option (B) $1, 2$: sum $=3 \neq -3$. This option fails.
Option (C) $0, -1$: product $=0 \neq 2$. This option fails.
Option (D) $0, 1$: product $=0 \neq 2$. This option fails.

Step 4: Confirm with the characteristic equation.
Since the sum is $-3$ and the product is $2$, the eigenvalues are roots of $\lambda^2 - (\text{sum})\lambda + \text{product} = 0$, that is $\lambda^2 + 3\lambda + 2 = 0$, which factors as $(\lambda+1)(\lambda+2)=0$, giving $\lambda = -1, -2$. This matches option (A) exactly.

Final Answer:
The eigenvalues are $-1$ and $-2$, option (A).
\[ \boxed{-1 \text{ and } -2} \]