Question

If matrix A = 3 4 1 -1 then show that Apower n = 1+2n -4n n 1-2n

Answer 1

Given :

$A=\left[\begin{array}{ll}3 & -4 \\1 & -1\end{array}\right]$

We have o show that $A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\n & 1-2 n\end{array}\right]$

To solve this matrix we have to use the concept of mathematical induction .

$P(n) \text { : If } A=\left[\begin{array}{cc}3 & -4 \\1 & -1\end{array}\right] \text {, then } A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\n & 1-2 n\end{array}\right], n \in N$

Let n = 1.

$\text { L.H.S }=A^{1}=A=\left[\begin{array}{ll}3 & -4 \\1 & -1\end{array}\right]$

$\text { R.H.S }=\left[\begin{array}{cc}1+2(1) & -4(1) \\1 & 1-2(1)\end{array}\right]$

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

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

$\text { L.H.S = R.H.S }$

$\therefore \mathrm{P}(\mathrm{n}) \text { is true for } \mathrm{n}=1$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\begin{gathered}A=\left[\begin{array}{ll}3 & -4 \\1 & -1\end{array}\right]\end{gathered}$

In order to show that,

$\\ \rm :\longmapsto\:\begin{gathered}A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\n & 1-2 n\end{array}\right]\end{gathered}$

We use the concept of Principal of Mathematical Induction

Step :- 1 For n = 1, we get

$\\ \rm :\longmapsto\:\begin{gathered}A^{1}=\left[\begin{array}{cc}1+2 & -4 \\1 & 1-2 \end{array}\right]\end{gathered}$

$\sf\implies \: \begin{gathered}A=\left[\begin{array}{ll}3 & -4 \\1 & -1\end{array}\right]\end{gathered}$

$\bf\implies \: {A}^{n} \: is \: true \: for \: n = 1$

Step :- 2 For n = k, Assume that it is true

$\rm :\longmapsto\:\begin{gathered}A^{k}=\left[\begin{array}{cc}1+2k & -4k \\k & 1-2k\end{array}\right]\end{gathered}$

Step :- 3 We have to show that it is true for n = k + 1

$\rm :\longmapsto\:\begin{gathered}A^{k + 1}=\left[\begin{array}{cc}3+2k & -4k - 4 \\k + 1 & - 1-2k\end{array}\right]\end{gathered}$

Consider,

$\rm :\longmapsto\: {A}^{k + 1}$

$\rm \: = \: {A}^{k} \times A$

$\rm \: = \: \begin{gathered}\left[\begin{array}{cc}1+2k & -4k \\k & 1-2k\end{array}\right]\end{gathered} \times \begin{gathered}\left[\begin{array}{ll}3 & -4 \\1 & -1\end{array}\right]\end{gathered}$

$\rm \: = \: \begin{gathered}\left[\begin{array}{cc}3+6k - 4k & -4 - 8k + 4k \\3k + 1 - 2k & - 4k - 1 + 2k\end{array}\right]\end{gathered}$

$\rm \: = \: \:\begin{gathered}\left[\begin{array}{cc}3+2k & -4k - 4 \\k + 1 & - 1-2k\end{array}\right]\end{gathered}$

$\bf\implies \: {A}^{n} \: is \: true \: for \: n = k + 1$

Hence,

By the Process of Principal of Mathematical Induction

$\\ \rm :\longmapsto\:\begin{gathered}A^{n}=\left[\begin{array}{cc}1+2 n & -4 n \\n & 1-2 n\end{array}\right]\end{gathered}$