If $I$ is the identity matrix of order 2 and $A=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array} ight]$, then for $n \geq 1$, mathematical induction gives

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If $I$ is the identity matrix of order 2 and $A=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$, then for $n \geq 1$, mathematical induction gives

Aicharya · Accepted Answer

Given $\quad A=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$ Now, $\quad A^2=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$ $ =\left[\begin{array}{ll} 1+0 & 1+1 \ 0+0 & 0+1 \end{array}ight]=\left[\begin{array}{ll} 1 & 2 \ 0 & 1 \end{array}ight] $ Similarly, $ A^3=\left[\begin{array}{ll} 1 & 3 \ 0 & 1 \end{array}ight] $ $ A^n=\left[\begin{array}{ll} 1 & n \ 0 & 1 \end{array}ight] $ We have, $n A-(n-1) I$ $ \begin{aligned} & =\left[\begin{array}{ll} n & n \ 0 & n \end{array}ight]-\left[\begin{array}{cc} n-1 & 0 \ 0 & n-1 \end{array}ight] \ & =\left[\begin{array}{ll} 1 & n \ 0 & 1 \end{array}ight]=A^n \ A^n & =n A-(n-1) / 	ext { is true } \end{aligned} $

If $I$ is the identity matrix of order 2 and $A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$,…

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