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If A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right] then prove that A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right] where n is any positive integer.
CBSE | Class 12 | Maths | Matrices | NCERT
If A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right] then prove that A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right] where n is any positive integer.

Useing PMI:

Result is true for n=1.

A^{1}=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]

Assume that result is true for n=k

\mathrm{A}^{k}=\left[\begin{array}{cr}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]

Prove that, result is true for n=k+1

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.:

A^{k+1}=A^{k} A

=\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ (k+1) & 1-2(k+1)\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right]

Using result from step 2 .

=\left[\begin{array}{ll}3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k\end{array}\right]

=\left[\begin{array}{ll}3+2 k & -4-4 k \\ 1+k & -1-2 k\end{array}\right]

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

=\mathrm{R} . \mathrm{H.S}

Thus, result is true.

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

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