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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