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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