If the matrix $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 3 & 0 & -1 \end{bmatrix}$ satisfies 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 $\alpha$ and $\beta$, then $\beta - \alpha$ is equal to ______.

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Aicharya · Accepted Answer

$A=\begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 3 & 0 & -1 \end{bmatrix}$ $A^2=\begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$, $A^3=\begin{bmatrix} 1 & 0 & 0 \ 0 & 8 & 0 \ 3 & 0 & -1 \end{bmatrix}$ $A^4=\begin{bmatrix} 1 & 0 & 0 \ 0 & 16 & 0 \ 0 & 0 & 1 \end{bmatrix}$ Hence, $A^{20}=\begin{bmatrix} 1 & 0 & 0 \ 0 & 2^{20} & 0 \ 0 & 0 & 1 \end{bmatrix}$, $A^{19}=\begin{bmatrix} 1 & 0 & 0 \ 0 & 2^{19} & 0 \ 3 & 0 & -1 \end{bmatrix}$ So, $A^{20}+\alpha A^{19}+\beta A=\begin{bmatrix} 1+\alpha+\beta & 0 & 0 \ 0 & 2^{20}+\alpha 2^{19}+2\beta & 0 \ 3\alpha+3\beta & 0 & 1-\alpha-\beta \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 1 \end{bmatrix}$ Therefore, $\alpha+\beta=0$ and $2^{20}+2^{19}\alpha-2\alpha=4$ $\Rightarrow \alpha=\frac{4(1-2^{18})}{2(2^{18}-1)}=-2$ Hence, $\beta=2$ So, $\beta-\alpha=4$

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$ satisfies the…

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