Q43 / 160 · TS EAMCET 2013 · Mathematics
If $I$ is the identity matrix of order 2 and $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$, then for $n \geq 1$, mathematical induction gives
- Correct answerA$A^n=n A-(n-1) I$
- B$A^n=n A+(n-1) I$
- C$A^n=2^n A-(n+1) I$
- D$A^n=2^{n-1} A-(n-1) I$
Solution
Given $\quad A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
Now, $\quad A^2=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$
$$
=\left[\begin{array}{ll}
1+0 & 1+1 \\
0+0 & 0+1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]
$$
Similarly,
$$
A^3=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]
$$
$$
A^n=\left[\begin{array}{ll}
1 & n \\
0 & 1
\end{array}\right]
$$
We have, $n A-(n-1) I$
$$
\begin{aligned}
& =\left[\begin{array}{ll}
n & n \\
0 & n
\end{array}\right]-\left[\begin{array}{cc}
n-1 & 0 \\
0 & n-1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & n \\
0 & 1
\end{array}\right]=A^n \\
A^n & =n A-(n-1) / \text { is true }
\end{aligned}
$$
Hence, the correct answer is option: 'A'.