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  • If A = [3 - 4 1 - 1], then prove that A^n [ 1 + 2n - 4n n 1 - 2n], where n is any positive integer.

Q3.    If A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}, then prove that A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix}, where n is any positive integer.

Answers (1)

best_answer

Given : 

A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}

To prove:

              A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix}

For n=1, we have  

                             A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix}= \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A

Thus, result is true for n=1.

Now, take result is true for  n=k, 

                         A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}

                          

For, n=k+1,

                  A^{K+1}=A.A^K

                               = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}

                                 =\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}

                               =\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}

                             =\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}

                           =\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}

                                 

Thus, the result is true for n=k+1.

Hence, we have   A^n = \begin{bmatrix} 1+2n & -4\n\\ n & 1-2n \end{bmatrix},    where A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}.

Posted by

seema garhwal

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