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If I is the identify matrix of order 2 and $A= \begin {bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, then for $ n \geq 1$, mathematical induction gives

\[\begin {array} {1 1} (a)\; A^n=nA-(n-1)I & \quad (b)\;A^n=nA+(n-1)I \\ (c)\;A^n=2^n A -(n+1)I & \quad (d)\;A^n=2^{n-1}A-(n-1)I \end {array}\]
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$(a)\; A^n=nA-(n-1)I$
answered Nov 7, 2013 by pady_1

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