Question

Let $A$ be a square matrix of order $n\times n$. A constant $\lambda$ is said to be characteristic root of $A$ if there exists a $n\times 1$ matrix $X$ such that $AX=\lambda X$

If $0$ is a characteristic root of $A$, then :

• A. $A$ is non-singular
• B. $A$ is singular
• C. $A=0$
• D. $A=I_n$

Answer 1

Answer: (B)

$A$ is singular

Since $X\ne 0$ is such that $(A-\lambda I)X=0,|A-\lambda I|=0\iff A-\lambda I$ is singular. If $A-\lambda I$ is non-singular the then equation $(A-\lambda I)X=0\Rightarrow X=0$
If $\lambda =0$, we get $|A|=0\Rightarrow A$ is singular.