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Question

Let $$A = \left[ {\matrix{   1 & 0  \cr    1 & 1  \cr 
 } } \right],$$ and $$I = \left[ {\matrix{   1 & 0  \cr    0 & 1  \cr 
 } } \right]$$ then prove that $${A^n} = nA - (n - 1)I,n \ge 1.$$


Solution

$$A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\qquad I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
if $$n=1$$
$$A=1\cdot A-(1-1)\cdot I$$
or $$A=A$$
if $$n=2$$
$${ A }^{ 2 }=A\times A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\\ \quad \quad \quad \quad \quad \quad =\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ 2\cdot A=\begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}\qquad (n-1)I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \therefore { A }^{ 2 }=2\cdot A-(2-1)I$$
Let's assume it is true for $$n=m$$
$${ A }^{ m }=m\cdot A-(m-1)\cdot I=\begin{bmatrix} 1 & 0 \\ m & 1 \end{bmatrix}\\ { A }^{ m }\cdot A=\begin{bmatrix} 1 & 0 \\ m & 1 \end{bmatrix}\cdot \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\\ \quad \quad \quad =\begin{bmatrix} 1 & 0 \\ m+1 & 1 \end{bmatrix}\\ (m+1)\cdot A-(m+1-1)I\\ =\begin{bmatrix} m+1 & 0 \\ m+1 & m+1 \end{bmatrix}-\begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}\\ =\begin{bmatrix} 1 & 0 \\ m+1 & 1 \end{bmatrix}$$
$$\therefore$$ By induction it is proved that 
$${ A }^{ n }=nA-(n-1)A$$ for $$n\ge 1$$

Mathematics

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