If then prove that where is any positive integer.

Useing PMI:

Result is true for .

Assume that result is true for

Prove that, result is true for

That is,

Ak+1=1+2(k+1)–4(k+1)(k+1)1–2(k+1)
A^{k+1}=\left[\begin{array}{cc}
1+2(k+1) & -4(k+1) \\
(k+1) & 1-2(k+1)
\end{array}\right]

L.H.S.:

Using result from step

Thus, result is true.

Therefore, By Mathematical Induction result is true for all positive integers.