# If     then which one of the following holds for all $\dpi{100} n\geq 1$ , by the principle of mathematical induction Option 1) $A^{n}=2^{n-1}A-(n-1)I\; \;$ Option 2) $\; A^{n}=nA-(n-1)I\; \;$ Option 3) $\; A^{n}=2^{n-1}A+(n-1)I\;$ Option 4) $\; A^{n}=nA+(n-1)I$

As we learnt in

Multiplication of matrices -

-

$A=\begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}$

$I=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$A^{n}= nA- (n-1)I$

$A= \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}$

$A^{2}= \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix} \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ 2& 1 \end{bmatrix}$

$A^{3}= \begin{bmatrix} 1 &0 \\ 2& 1 \end{bmatrix} \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ 3& 1 \end{bmatrix}$

$A^{n}= \begin{bmatrix} 1 &0 \\ n& 1 \end{bmatrix}$

now $n(A)= \begin{bmatrix} n &0 \\ n& n \end{bmatrix}$

$(n-1)I= (n-1)\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$$= \begin{bmatrix} n-1 &0 \\ 0& n-1 \end{bmatrix}$

$\therefore \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}$

Option 1)

$A^{n}=2^{n-1}A-(n-1)I\; \;$

Incorrect option

Option 2)

$\; A^{n}=nA-(n-1)I\; \;$

Correct option

Option 3)

$\; A^{n}=2^{n-1}A+(n-1)I\;$

Incorrect option

Option 4)

$\; A^{n}=nA+(n-1)I$

Incorrect option

N

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