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If the matrix A =\begin{bmatrix} 1 &0 &0 \\ 0&2 &0 \\ 3&0 &-1 \end{bmatrix} satisfy the equation A^{20 }+\alpha A^{19} +\beta A = \begin{bmatrix} 1 &0 &0 \\ 0 &4 &0 \\ 0& 0 &1 \end{bmatrix} for some real numbers \alpha\text{ and } \beta, then \beta- \alpha is equal to _______
Option: 1 3
Option: 2 2
Option: 3 1
Option: 4 4

Answers (1)

best_answer

Given that 

A=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{array}\right]

\mathrm{A}^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array}\right], \mathrm{A}^{3}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1 \end{array}\right]

\mathrm{A}^{4}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1 \end{array}\right]

Hence,

\mathrm{A}^{20}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{array}\right], \mathrm{A}^{19}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1 \end{array}\right]

\\\text { So } A^{20}+\alpha A^{19}+\beta A=\left[\begin{array}{ccc} 1+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha \cdot 2^{19}+2 \beta & 0 \\ 3 \alpha+3 \beta & 0 & 1-\alpha-\beta \end{array}\right]\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}\quad\quad\quad\;\;\;\;\;\;\;\;\;=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array}\right]

\\\text { Therefore } \alpha+\beta=0 \text { and } 2^{20}+2^{19} \alpha-2 \alpha=4 \\ \Rightarrow \alpha=\frac{4\left(1-2^{18}\right)}{2\left(2^{18}-1\right)}=-2 \\ \text { hence } \beta=2 \\ \text { so }(\beta-\alpha)=4

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himanshu.meshram

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