Question

Let A be a square matrix of order n × n. A constant λ is said to be the characteristic root of A if there exists a n × 1 matrix X such that AX = λX. If λ is a characteristic root of A then:
A. A – λI = 0
B. A – λI is singular
C. A – λI is non-singular
D. None of these

Answer 1

Hint: Here, A is a square matrix and λ is said to be the characteristic root of A. Hence, A ≠ 0. Solve the equation given assuming A is not equal to 0. Find the condition of the equation and compare it with given options and choose the correct option.

Complete step by step solution:
In these types of questions, always consider the given equation carefully. Matrix operations and algebraic operations are similar sometimes but be careful while doing matrix operation, because matrix multiplication is not commutative while in algebra multiplication is commutative. While solving matrix linear operations carefully shift terms. Matrix 0 is not considered as 0 only but it is considered as zero matrix.
Singular Matrix: A square matrix A is said to be singular matrix if |A| = 0.
Since, A is a square matrix of order n × n, then X ≠ 0.
Given that AX = λX
⇒ (A − λI) X = 0 ⇒ ∣A − λI∣ = 0 ⇔ A − λI is singular.
Hence, the correct option is (B).
So, the correct answer is “Option B”.

Note: Non-singular Matrix: A square matrix A is said to be non-singular matrix if |A| ≠ 0.
Here, in options two options (A) and (B) are similar but do not get confused with the two. In option (A), A – λI = 0 this means the value of matrix (A – λI) is zero or zero matrix. But in option (B), A – λI = 0 is singular means the determinant of (A – λI) is zero. So, the correct option is (B).