If     then which one of the following holds for all 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

 

Answers (2)
V Vakul

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 neha

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