Question

If A = \(\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}\) then \((aI + bA)^n\) is (where I is the identity matrix of order 2)

• A. \(a^2I + a^{n-1}b \cdot A\)
• B. \(a^n I + na^n b \cdot A\)
• C. \(a^nI + n \cdot a^{n-1} b \cdot A\)
• D. \(a^nI + b^nA\)

Answer 1

The Correct Option is C

We are given: \[ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Note that: \[ A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0 \] So, since \( A^2 = 0 \), the matrix \( A \) is nilpotent of index 2. Now, expand \( (aI + bA)^n \) using the binomial theorem: \[ (aI + bA)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (bA)^k \] But since \( A^2 = 0 \), all terms with \( k \geq 2 \) vanish. So, we keep only terms for \( k = 0 \) and \( k = 1 \): \[ (aI + bA)^n = a^n I + n a^{n-1} b A \] Correct answer: \( a^n I + n \cdot a^{n-1} b \cdot A \) 

Answer 2

Given A = \(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) and I = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), we want to find (aI + bA)ⁿ.

First, note that A² = \(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\). Therefore A^k = 0 for k ≥ 2.

aI + bA = a\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) + b\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) = \(\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\)

Now, let's examine (aI + bA)ⁿ for small values of n:

n = 1: (aI + bA)¹ = \(\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\) = aI + bA

n = 2: (aI + bA)² = \(\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\)\(\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\) = \(\begin{bmatrix} a^2 & 2ab \\ 0 & a^2 \end{bmatrix}\) = a²I + 2abA

n = 3: (aI + bA)³ = (aI + bA)²(aI + bA) = \(\begin{bmatrix} a^2 & 2ab \\ 0 & a^2 \end{bmatrix}\)\(\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\) = \(\begin{bmatrix} a^3 & a^2b + 2a^2b \\ 0 & a^3 \end{bmatrix}\) = \(\begin{bmatrix} a^3 & 3a^2b \\ 0 & a^3 \end{bmatrix}\) = a³I + 3a²bA

From these examples, we can see a pattern emerging:

(aI + bA)ⁿ = aⁿI + n * a^(n-1) * b * A

This can be proved through induction as well.

Answer:

\(a^nI + n \cdot a^{n-1} b \cdot A\)