Question

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

Solution

## It is given that $$\displaystyle A=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$To prove: $$\displaystyle P\left( n \right) :{ A }^{ n }=\begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix},n\in N$$We shall prove the result by using the principle of mathematical induction .For $$\displaystyle n=1$$, we have:$$\displaystyle P\left( 1 \right) :{ A }^{ 1 }=\begin{bmatrix} 1+2 & -4 \\ 1 & 1-2 \end{bmatrix}=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}=A$$Therefore, the result is true for $$\displaystyle n=1$$.Let the result be true for $$\displaystyle n=k$$.That is , $$\displaystyle P\left( k \right) :{ A }^{ k }=\begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix},n\in N$$Now, we prove that the result is true for $$\displaystyle n=k+1$$Consider$$\displaystyle { A }^{ k+1 }={ A }^{ k }.A$$$$\displaystyle =\begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$$$\displaystyle =\begin{bmatrix} 3\left( 1+2k \right) -4k & -4\left( 1+2k \right) +4k \\ 3k+1-2k & -4k-1\left( 1-2k \right) \end{bmatrix}$$$$\displaystyle =\begin{bmatrix} 3+6k-4k & -4-4k \\ 3k+1-2k & -4k-1+2k \end{bmatrix}$$$$\displaystyle =\begin{bmatrix} 3+2k & -4-4k \\ 1+k & -1-2k \end{bmatrix}$$$$\displaystyle =\begin{bmatrix} 1+2\left( k+1 \right) & -4\left( k+1 \right) \\ 1+k & 1-2\left( k+1 \right) \end{bmatrix}$$Therefore, the result is true for $$\displaystyle n=k+1$$.Thus, by the principle of mathematical induction , we have:$$\displaystyle { A }^{ n }=\begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix},n\in N$$MathematicsNCERTStandard XII

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