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If $A$ is a matrix $\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$, then…
If $A$ is a matrix $\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$, then $A^n=\left[\begin{array}{ll}1 & 0 \\ n & 1\end{array}\right]$, $\forall n \in N$
Not true for n = 3 Not true for n = 2 True for n = 3 Not true for n = 1 Solution
Given, If $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$, then $A^n=\left[\begin{array}{ll}1 & 0 \\ n & 1\end{array}\right], \forall n \in N$
We will generalise the exponent of $A$.
Let $n=2$
$
\begin{aligned}
\therefore A^2 & =\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \cdot\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 \cdot 1+0 \cdot 1 & 1 \cdot 0+0 \cdot 1 \\
1 \cdot 1+1 \cdot 1 & 1 \cdot 0+1 \cdot 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]
\end{aligned}
$
Again, let $n=3$
$
\therefore A^3=A^2 \cdot A=\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right] \cdot\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
2+1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]
$
Similarly, $A^4=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
Hence, $A^n=\left[\begin{array}{ll}1 & 0 \\ n & 1\end{array}\right]$ is true for $n=3$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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