I'm having trouble with some basic linear algebra. My best searching doesn't exactly come up with the solution I'm looking for. My problem is as follows.

$$ \begin{bmatrix} y_1 \\ y_2 \\ \end{bmatrix}=A \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} ,~~\text{and}~~ \begin{bmatrix} y_2 \\ y_3 \\ \end{bmatrix}=B \begin{bmatrix} x_3 \\ x_2 \\ \end{bmatrix} $$

where $y_1$ and $x_2$ are known. I see that I can rewrite the systems of equations as

$$ \begin{bmatrix} x_1 \\ y_2 \\ \end{bmatrix}=A' \begin{bmatrix} y_1 \\ x_2 \\ \end{bmatrix} ~\text{where}~A'= \frac{1}{a_{11}} \begin{bmatrix} 1 & -a_{12}\\ b_{21} & \det{A}\\ \end{bmatrix} \tag{1}\label{eq1} $$ and similarly $$ \begin{bmatrix} x_3 \\ y_3 \\ \end{bmatrix}=B' \begin{bmatrix} y_2 \\ x_2 \\ \end{bmatrix} ~\text{where}~B'= \frac{1}{b_{11}} \begin{bmatrix} 1 & -b_{12}\\ b_{21} & \det{B}\\ \end{bmatrix} \tag{2}\label{eq2} $$

This lets me solve by evaluating \ref{eq1} then plug $y_2$ into \ref{eq2}. However, I'd like to avoid the "plugging" in part. How can I arrive at that solution through matrix operations?

In other words, how do I get to something like this $$ \begin{bmatrix} x_3 \\ y_3 \\ \end{bmatrix}=C \begin{bmatrix} y_1 \\ x_2 \\ \end{bmatrix} $$



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.