Question

Let matrix $M=\begin{pmatrix} 4 & x \\ 6 & 9\end{pmatrix}$. If $\det(M)=0$, then

• A. $M$ is symmetric.
• B. $M$ is invertible.
• C. One eigenvalue is 13.
• D. Its eigenvectors are orthogonal.

Hint:

A zero determinant always means one eigenvalue is zero; the other equals the trace.

Answer 1

The Correct Option is A, C, D

Step 1: Use determinant condition.
$\det(M) = 4\cdot 9 - 6x = 36 - 6x = 0$ $\Rightarrow x = 6.$

Step 2: Write the matrix.
$M = \begin{pmatrix}4 & 6 \\ 6 & 9\end{pmatrix}.$

Step 3: Compute eigenvalues.
Solve $\det(M - \lambda I)=0$: $\big(4-\lambda)(9-\lambda) - 36 = 0$ $\Rightarrow \lambda^2 - 13\lambda = 0$ $\Rightarrow \lambda(\lambda - 13)=0.$
Eigenvalues are $0$ and $13$.

Step 4: Conclusion.
One eigenvalue is 13 ⇒ option (C).