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Question

If A = $$\begin{vmatrix}
1 &0 \\
 1& 1
\end{vmatrix}$$B and I =$$\begin{vmatrix}
1 &0 \\
0& 1
\end{vmatrix}$$ ,then which one of the following holds for all n $$\geq $$ 1, by
the principle of mathematical indunction 


A
An=nA(n1)l
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B
An=2n1A(n1)l
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C
An=nA+(n1)l
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D
An=2n1A+(n1)l
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Solution

The correct option is A $$A^{n}=nA-(n-1)l$$
$$A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\\ { A }^{ 2 }=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ { A }^{ 3 }=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}\\ \therefore { A }^{ n }=\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$$

and $$nA=n\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}$$

$$\left( n-1 \right) I=\left( n-1 \right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}\\ \therefore nA-\left( n-1 \right) I=\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}-\begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}={ A }^{ n }$$

Mathematics

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