If $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$, show that $A^k = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix}$, where $k$ is any positive integer.
Step-by-Step Solution
Key Concept: General
We have,<br>$\Rightarrow A^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 1+2\times2 & -4\times2 \\ 2 & 1-2\times2 \end{bmatrix}$ and<br>$\Rightarrow A^3 = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 7 & -12 \\ 3 & -5 \end{bmatrix} = \begin{bmatrix} 1+2\times3 & -4\times3 \\ 3 & 1-2\times3 \end{bmatrix}$<br>Thus, it is true for indices 2 and 3. Now assume<br>$\Rightarrow A^k = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix}$<br>Then, $A^{k+1} = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3+2k & -4(k+1) \\ k+1 & -1-2k \end{bmatrix} = \begin{bmatrix} 1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1) \end{bmatrix}$<br>Thus, if the law is true for $A^k$, it is also true for $A^{k+1}$. But it is true for $k = 2, 3$ etc. Hence, by induction, the required result follows.
Correct Answer: A