Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade 12
Question:
If $A$ and $B$ are square matrices of same order such that they commute then:
A^m & B^n commute, m,n \in \mathbb{N}
(A+B)^n = \sum C_0 A^n + C_1 A^{n-1}B + \ldots + nC_n B^n, n \in \mathbb{N}
A - \lambda I, B + \mu I commute \forall \lambda, \mu \in \mathbb{R}
A + \lambda I, \mu I - B commute \forall \lambda, \mu \in \mathbb{R}
Step-by-Step Solution
Key Concept: When matrices commute ($AB = BA$), scalar multiples and shifted versions preserve commutativity, and binomial expansions become valid.
Given that $AB = BA$, we verify each option: **(1)** If $AB = BA$, then $A^m B^n = B^n A^m$ for any $m,n \in \mathbb{N}$ (proven by induction), so they commute. **(2)** The binomial expansion $(A+B)^n = \sum_{k=0}^{n} C_k A^{n-k}B^k$ holds when matrices commute, since we can rearrange terms freely. **(3)** Since $AB = BA$, we have $(A-\lambda I)B = AB - \lambda B = BA - \lambda B = B(A-\lambda I)$, and similarly $(B+\mu I)$ commutes with $A$. **(4)** Similarly, $(A+\lambda I)(\mu I - B) = \mu A - AB + \mu \lambda I - \lambda B = \mu A - BA + \mu \lambda I - \lambda B = (\mu I - B)(A + \lambda I)$. All four statements are universally true for commuting matrices.
Correct Answer: 1,2,3,4