Question:
The matrix $\left[\begin{matrix}1& 4\\ 3& 2\end{matrix}\right]$ is
The matrix $\left[\begin{matrix}1& 4\\ 3& 2\end{matrix}\right]$ is
Show Hint
"Distinct Eigenvalues $\implies$ Diagonalizable." It's the most common sufficient condition.
Updated On: Jul 3, 2026
- symmetric
- diagonalizable
- skew symmetric
- singular
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The Correct Option is B
Solution and Explanation
Step 1: Concept
A matrix is diagonalizable if it has a complete set of linearly independent eigenvectors, which is guaranteed if all its eigenvalues are distinct.
Step 2: Meaning
Calculate the eigenvalues using the characteristic equation $\det(A - \lambda I) = 0$: $(1-\lambda)(2-\lambda) - 12 = 0$.
Step 3: Analysis
$\lambda^2 - 3\lambda + 2 - 12 = 0 \Rightarrow \lambda^2 - 3\lambda - 10 = 0$. Factoring gives $(\lambda - 5)(\lambda + 2) = 0$, so $\lambda_1 = 5, \lambda_2 = -2$.
Step 4: Conclusion
Since the matrix has two distinct eigenvalues for a $2 \times 2$ size, it must be diagonalizable. Final Answer: (B)
A matrix is diagonalizable if it has a complete set of linearly independent eigenvectors, which is guaranteed if all its eigenvalues are distinct.
Step 2: Meaning
Calculate the eigenvalues using the characteristic equation $\det(A - \lambda I) = 0$: $(1-\lambda)(2-\lambda) - 12 = 0$.
Step 3: Analysis
$\lambda^2 - 3\lambda + 2 - 12 = 0 \Rightarrow \lambda^2 - 3\lambda - 10 = 0$. Factoring gives $(\lambda - 5)(\lambda + 2) = 0$, so $\lambda_1 = 5, \lambda_2 = -2$.
Step 4: Conclusion
Since the matrix has two distinct eigenvalues for a $2 \times 2$ size, it must be diagonalizable. Final Answer: (B)
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