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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:
$$\displaystyle { A }^{ n }=\begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix},n\in N$$

Mathematics
NCERT
Standard XII

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