Question

Consider the matrix \[A = \begin{bmatrix} 5 & -4 \\ k & -1 \end{bmatrix},\] 
where \(k\) is a constant. If \(\det(A) = 3\), then the ratio of the largest eigenvalue of \(A\) to \(k\) is ___________ (rounded off to 1 decimal place).
 

Hint:

For \(2\times2\) matrices, you often don't need the full characteristic expansion: the sum and product of eigenvalues are the trace and determinant.

Answer 1

Step 1: Use the determinant condition to find \(k\).
\[ \det A = 5(-1) - (-4)k = -5 + 4k = 3 \;\Rightarrow\; 4k=8 \;\Rightarrow\; k=2. \] 

Step 2: Find the eigenvalues using trace and determinant.
For any \(2\times2\) matrix, the eigenvalues \(\lambda_{1,2}\) satisfy \[ \lambda_1+\lambda_2=\operatorname{tr}A=5+(-1)=4,\qquad \lambda_1\lambda_2=\det A=3. \] Hence \(\lambda\) solves \( \lambda^2-4\lambda+3=0\Rightarrow(\lambda-1)(\lambda-3)=0\).
So the eigenvalues are \(1\) and \(3\); the largest is \(3\). 

Step 3: Compute the requested ratio.
\[ \frac{\lambda_{\max}}{k}=\frac{3}{2}=1.5. \] \[ \boxed{1.5} \]