Question

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

Answer 1

Step1. Calculate the value of ${\mathrm{A}}^{19}\mathrm{and}{\mathrm{A}}^{20}$:

$\begin{array}{rcl}A& =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\\ A^2& =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\times \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 2^2& 0\\ 0& 0& 1\end{array}\right]\\ A^3& =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\times \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\times \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 8& 0\\ 3& 0& -1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 2^3& 0\\ 3& 0& -1\end{array}\right]\\ A^4& =& A^2\times A^2\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\times \left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 16& 0\\ 0& 0& 1\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 2^4& 0\\ 0& 0& 1\end{array}\right]\\ & & \end{array}$

Therefore,

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

Step2. Calculate the value of $\beta -\alpha$:

$\begin{array}{rcl}{A}^{20}+ɑA^{19}+\beta A& =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\\ & \Rightarrow & \left[\begin{array}{ccc}1& 0& 0\\ 0& 2^{20}& 0\\ 0& 0& 1\end{array}\right]+\alpha \left[\begin{array}{ccc}1& 0& 0\\ 0& 2^{19}& 0\\ 3& 0& -1\end{array}\right]+\beta \left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 3& 0& -1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\\ & \Rightarrow & \left[\begin{array}{ccc}1+\alpha +\beta & 0& 0\\ 0& 2^{20}+\alpha 2^{19}+2\beta & 0\\ 3\alpha +3\beta & 0& 1-\alpha -\beta \end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& 0\\ 0& 0& 1\end{array}\right]\\ & & \end{array}$

Therefore,

$$\begin{gathered} \Rightarrow 1+\alpha +\beta =1,2^{20}+\alpha 2^{19}+2\beta =4,3\alpha +3\beta =0,1-\alpha -\beta =1 \\ \Rightarrow \alpha +\beta =0 \\ \Rightarrow \alpha =-\beta \end{gathered}$$

Step 3. Replace $\alpha =-\beta$ in given equation:

$2^{20}+\alpha 2^{19}+2\beta =4$

$\Rightarrow$$2^{20}+\left(-\beta \right)2^{19}+2\beta =4$

$\Rightarrow$ $\mathrm{\beta }=\frac{4-2^{20}}{2-2^{19}}$

$\Rightarrow$ $\mathrm{\beta }=\frac{-1048572}{-524286}$

$\Rightarrow$ $\beta =2$

$\Rightarrow$ $\alpha =-2$

$\begin{array}{rcl}\therefore \beta -\alpha & =& 2-\left(-2\right)\\ & =& 4\end{array}$

Hence, the value of $\mathbf{\beta }\mathbf{}\mathbf{-}\mathit{\alpha }\mathbf{=}\mathbf{4}$ is the answer