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Question

$$ A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \end{bmatrix} $$ then prove that $$ A^{ n }=\begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix} $$ where n is any positive integer.


Solution

let $$ P(n) $$ be the statement
$$ P(n) \Rightarrow \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix} $$
$$ P(1)  = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}=A $$
$$ \therefore P(1) $$ is true.
Let $$ P(k) $$ is true.
$$ P(k) = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix}=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} $$
$$ = \begin{bmatrix} (1+2k)3+(-4k)1 & (1+2k)(-4)\quad +(-4k)(-1) \\ k.3+(1-2k)1 & k(-4)+(1-2k)(-1)\quad  \end{bmatrix} \begin{bmatrix} 3+2k & -4-4k \\ k+1 & -1-2k \end{bmatrix}=\begin{bmatrix} 1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\quad \quad  \end{bmatrix} $$
$$ \therefore P ( k + 1) $$ is true.
$$ P(n) $$ is true for all the values of $$ +ve n $$

Mathematics

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