Math, asked by Razlan53931, 11 months ago

यदि  A = \begin{bmatrix}  1 & 1 & 1 \\  1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix},तो सिद्ध कीजिए कि  An = \begin{bmatrix}  3n-1 & 3n-1 & 3n-1 \\  3n-1 & 3n-1 & 3n-1 \\ 3n-1 & 3n-1 & 3n-1 \end{bmatrix}, n ∈ N

Answers

Answered by amitnrw
0

Given :   A =  \begin{bmatrix}  1 & 1 & 1 \\  1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}    n ∈ N    

To find :   सिद्ध कीजिए कि Aⁿ = \begin{bmatrix}  3^{n-1} & 3^{n-1} & 3^{n-1} \\  3^{n-1} & 3^{n-1} & 3^{n-1} \\3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}

Solution:

Aⁿ = \begin{bmatrix}  3^{n-1} & 3^{n-1} & 3^{n-1} \\  3^{n-1} & 3^{n-1} & 3^{n-1} \\3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}

n = 1

LHS = A

RHS =    \begin{bmatrix}  1 & 1 & 1 \\  1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}  

LHS =  RHS

n = x

=>   Aˣ =  \begin{bmatrix}  3^{x-1} & 3^{x-1} & 3^{x-1} \\  3^{x-1} & 3^{x-1} & 3^{x-1} \\3^{x-1} & 3^{x-1} & 3^{x-1} \end{bmatrix}  

n = x + 1

=> LHS = Aˣ⁺¹

RHS  =    \begin{bmatrix}  3^{x } & 3^{x } & 3^{x } \\  3^{x } & 3^{x } & 3^{x } \\3^{x } & 3^{x } & 3^{x } \end{bmatrix}

LHS = Aˣ⁺¹ = Aˣ . A

= \begin{bmatrix}  3^{x-1} & 3^{x-1} & 3^{x-1} \\  3^{x-1} & 3^{x-1} & 3^{x-1} \\3^{x-1} & 3^{x-1} & 3^{x-1} \end{bmatrix}  \begin{bmatrix}  1 & 1 & 1 \\  1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}

=\begin{bmatrix}  3^{x-1} + 3^{x-1}  +  3^{x-1} & 3^{x-1} + 3^{x-1}  +  3^{x-1} & 3^{x-1} + 3^{x-1}  +  3^{x-1}  \\  3^{x-1} + 3^{x-1}  +  3^{x-1} & 3^{x-1} + 3^{x-1}  +  3^{x-1} & 3^{x-1} + 3^{x-1}  +  3^{x-1} \\ 3^{x-1} + 3^{x-1}  +  3^{x-1} & 3^{x-1} + 3^{x-1}  +  3^{x-1} & 3^{x-1} + 3^{x-1}  +  3^{x-1}\end{bmatrix}

= \begin{bmatrix}  3\times 3^{x-1} &  3\times 3^{x-1} &  3\times 3^{x-1} \\ 3\times 3^{x-1} &  3\times 3^{x-1} &  3\times 3^{x-1} \\3\times 3^{x-1} &  3\times 3^{x-1} &  3\times 3^{x-1} \end{bmatrix}

=  \begin{bmatrix}  3^{x } & 3^{x } & 3^{x } \\  3^{x } & 3^{x } & 3^{x } \\3^{x } & 3^{x } & 3^{x } \end{bmatrix}

=RHS

QED

इति सिद्धम

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