Is it possible to solve this system of equation, if not why not?

Question

Is it possible to solve this system of equation, if not why not?

$$\left[\begin{matrix}-\alpha_1&0\\0&-\alpha_2\\\end{matrix}\right]\left[\begin{matrix}1&{-\beta}_1\\{-\beta}_2&1\\\end{matrix}\right]^{-1}=\left[\begin{matrix}0.25&0\\0.25&0.25\\\end{matrix}\right]$$

To give some context, I am trying to simulate a system of equations based on a specified error term structure and this is a problem that I have come across.

What I have tried:

$$\left[\begin{matrix}-\alpha_1&0\\0&-\alpha_2\\\end{matrix}\right]\frac{1}{1+(\beta_1\beta_2)}\left[\begin{matrix}1&\beta_1\\\beta_2&1\\\end{matrix}\right]=\left[\begin{matrix}0.25&0\\0.25&0.25\\\end{matrix}\right]$$

Answer 1

From $\left[\begin{matrix}-\alpha_1&0\\0&-\alpha_2\\\end{matrix}\right]\frac{1}{1+(\beta_1\beta_2)}\left[\begin{matrix}1&\beta_1\\\beta_2&1\\\end{matrix}\right]=\left[\begin{matrix}0.25&0\\0.25&0.25\\\end{matrix}\right]$

We get

$\frac{1}{1+\beta_1\beta_2}\left[\begin{matrix}-\alpha_1&- \alpha_1 \beta_1\\\alpha_2 \beta_2&-\alpha_2\\\end{matrix}\right]=\left[\begin{matrix}0.25&0\\0.25&0.25\\\end{matrix}\right]$.

Since $ \alpha_1 \ne 0$, we derive from $\alpha_1 \beta_1=0$, that $ \beta_1=0.$

Hence we have

$\left[\begin{matrix}-\alpha_1&0\\\alpha_2 \beta_2&-\alpha_2\\\end{matrix}\right]=\left[\begin{matrix}0.25&0\\0.25&0.25\\\end{matrix}\right]$.

This yields $\alpha_1= \alpha_2 =-0.25$ and $ \beta_2=-1.$

Answer 2

A Hint, not a complete solution

$$ \left[\begin{matrix}-\alpha_1&0\\0&-\alpha_1\\\end{matrix}\right]\left[\begin{matrix}1&\beta_1\\\beta_1&1\\\end{matrix}\right]^{-1}=\left[\begin{matrix}0.25&0\\0.25&0.25\\\end{matrix}\right] $$ can be rewritten as $$ 4\left[\begin{matrix}-\alpha_1&0\\0&-\alpha_1\\\end{matrix}\right]=\left[\begin{matrix}1&0\\1&1\\\end{matrix}\right]\left[\begin{matrix}1&\beta_1\\\beta_1&1\\\end{matrix}\right] $$ (with the condition that $1 - \beta_1^2 \ne 0$, so that the rightmost matrix is invertible).

Evaluating the upper left hand corner of each side we get $$ -4 \alpha_1 = 1 \cdot 1 + 0 \cdot \beta_1 = 1. $$ So if there's a solution, then $$ \alpha_1 = -\frac{1}{4}. $$ I suspect you can look at the other four equations similarly, and rapidly discover whether there's a solution or not.