Question

Which of the following is one of the eigenvalues for the matrix given below?
\[ \begin{bmatrix} 3 & 4 \\ 4 & -3 \end{bmatrix} \]

• A. 1
• B. 3
• C. 5
• D. 7

Hint:

For a 2x2 matrix, eigenvalues satisfy sum = trace and product = determinant; here trace=0 and determinant=-25.

Answer 1

The Correct Option is C

Instead of writing out the full characteristic equation from scratch, use the trace and determinant shortcut for a $2\times2$ matrix.

1. Find the trace: the trace of $A = \begin{bmatrix}3 & 4\\4 & -3\end{bmatrix}$ is the sum of the diagonal entries: $\text{trace}(A) = 3 + (-3) = 0$. The trace also equals the sum of the two eigenvalues, so $\lambda_1 + \lambda_2 = 0$.
2. Find the determinant: $\det(A) = (3)(-3) - (4)(4) = -9 - 16 = -25$. The determinant equals the product of the eigenvalues, so $\lambda_1 \lambda_2 = -25$.
3. Solve the pair of equations: from $\lambda_1+\lambda_2=0$, write $\lambda_2 = -\lambda_1$. Substitute into $\lambda_1\lambda_2=-25$:
   $$\lambda_1(-\lambda_1) = -25 \implies -\lambda_1^2 = -25 \implies \lambda_1^2 = 25 \implies \lambda_1 = \pm5$$
4. Get both eigenvalues: if $\lambda_1 = 5$, then $\lambda_2 = -5$ (and vice versa). Either way, the pair of eigenvalues is $\{5, -5\}$.
5. Match to the options: among the given choices $1, 3, 5, 7$, only $5$ appears in the eigenvalue set $\{5,-5\}$.

Let's summarize:

• For a $2\times2$ matrix, the sum of eigenvalues equals the trace and the product equals the determinant, without needing to expand the full characteristic polynomial by hand.
• Here trace $=0$ and determinant $=-25$ give eigenvalues $5$ and $-5$.

So one of the eigenvalues is $5$, option (C).