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If $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then for all $n \in N$
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1787 Upvotes
Verified Answer
The correct answer is:
$A^n=n A-(n-1) t$
We have,
$$
A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \text { and } Y=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
$$
Now,
$$
\begin{aligned}
A^2 & =A \cdot A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
& =2\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=2 A-(2-1) Y
\end{aligned}
$$
$$
\begin{aligned}
& \text { Again, } A^3=A^2 \cdot A=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \\
& \quad=3\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=3 A-(3-1) Y \\
& \therefore A^n=n A-(n-1) Y
\end{aligned}
$$
$$
A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \text { and } Y=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
$$
Now,
$$
\begin{aligned}
A^2 & =A \cdot A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
& =2\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=2 A-(2-1) Y
\end{aligned}
$$
$$
\begin{aligned}
& \text { Again, } A^3=A^2 \cdot A=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \\
& \quad=3\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=3 A-(3-1) Y \\
& \therefore A^n=n A-(n-1) Y
\end{aligned}
$$
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