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, bythe principle of mathematical indunction

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

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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