Question

If the matrix $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]$ satisfies the equation $A^{20}+\alpha A^{19}+\beta A=A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]$ for some real numbers $\alpha$ and$\beta$, then $\beta -\alpha$ is equal to

Answer 1

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

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

$A^4=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 16& 0\\ 0& 0& 1\end{array}\right]$
$⋮⋮⋮$
$A^{19}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2^{19}& 0\\ 3& 0& -1\end{array}\right],A^{20}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 2^{20}& 0\\ 0& 0& 1\end{array}\right]$

L. H. S
$=A^{20}+\alpha A^{19}+\beta A=\left[\begin{array}{ccc}1+\alpha +\beta & 0& 0\\ 0& 2^{20}+\alpha 2^{19}+2\beta & 0\\ 3\alpha +3\beta & 0& 1-\alpha -\beta \end{array}\right]$
R.H.S
$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\Rightarrow \alpha +\beta =0$
and $2^{20}+\alpha 2^{19}+2\beta =4$
$\Rightarrow 2^{20}+\alpha (2^{19}-2)=4$
$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=-2\Rightarrow \beta =2$
$\therefore \beta -\alpha =4$