If A = [(3,-4),(1,-1)] then prove that A^n =[(1+2n,-4n),(n,1-2n)]

Question

If A =\(\begin{bmatrix} 3 & -4 \\[0.2em] 1& -1 \end{bmatrix}\) then prove that Aⁿ = \(\begin{bmatrix} 1+2n & -4n \\[0.2em] n & 1-2n \end{bmatrix}\)

Answer 1

Let the statement T(n) be: 

If A = \(\begin{bmatrix} 3 & -4 \\[0.2em] 1 & -1 \end{bmatrix}\) , then Aⁿ = \(\begin{bmatrix} 1+2n & -4n \\[0.2em] n & 1-2n \end{bmatrix}\) 

Basic Step: 

For n = 1, A= \(\begin{bmatrix} 3 & -4 \\[0.2em] 1 & -1 \end{bmatrix}\) 

A¹  = \(\begin{bmatrix} 1+2\times 1 & -4\times 1 \\[0.2em] 1 & 1-2\times 1 \end{bmatrix}\) = \(\begin{bmatrix} 3 & -4 \\[0.2em] 1 & -1 \end{bmatrix}\) 

⇒ T(1) is true

Induction Step: 

Assume T(k) to be true, i.e.,

If A =\(\begin{bmatrix} 3 & -4 \\[0.2em] 1 & -1 \end{bmatrix}\)  , then A^k = \(\begin{bmatrix} 1+2k & -4k \\[0.2em] k & 1-2k \end{bmatrix}\)

Now we need to show T(k + 1) is true. A^k = \(\begin{bmatrix} 1+2k & -4k \\[0.2em] k & 1-2k \end{bmatrix}\) 

∴ A^{k + 1} = A^k . A = \(\begin{bmatrix} 1+2k & -4k \\[0.2em] k & 1-2k \end{bmatrix}\)\(\begin{bmatrix} 3 & -4 \\[0.2em] 1 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 3+6k-4k & -4-8k+4k \\[0.2em] 3k+1-2k & -4k+2k-1 \end{bmatrix}\)

= \(\begin{bmatrix} 3+2k & -4-4k \\[0.2em] k+1 & -1-2k \end{bmatrix}\)  = \(\begin{bmatrix} 1+2(k+1) & -4(k+1) \\[0.2em] k+1 & 1-2(k+1) \end{bmatrix}\) 

⇒ T(k + 1) is true, whenever T(k) is true