Need help to get via the two given matrix-equations the desired (given) one.

Question

I need help getting from these two equations $$ \begin{bmatrix}Lz \\ y\end{bmatrix} = \frac{L}{2}\cdot \begin{bmatrix}1&n\\-1&n\end{bmatrix}\begin{bmatrix}P\\O\end{bmatrix} $$ $$ \begin{bmatrix}Lk \\ m\end{bmatrix} = \frac{L}{2}\cdot \begin{bmatrix}n&1\\-n&1\end{bmatrix}\begin{bmatrix}P\\O\end{bmatrix} $$ this $$ \begin{bmatrix}Lz \\ y\end{bmatrix} = \frac{1}{2n}\cdot \begin{bmatrix}1&n\\-1&n\end{bmatrix}\begin{bmatrix}1&-1\\n&n\end{bmatrix}\begin{bmatrix}Lk\\m\end{bmatrix} $$ I would appreciate any hints or full explanation, I'm new to matrices.

Answer 1

In general, if you have two vectorial equation for the vectors $\vec{x}$, $\vec{y}$ and $\vec{z}$ with the matrices $A$ and $B$ like the following: $$\vec{x} = A\vec{z}$$ $$\vec{y} = B\vec{z}$$ You can notice that for any matrix $C$ (of the correct dimensions) the following is true: $$C\vec{y} = CB\vec{z}$$ Now, if $B$ is invertible, then you can simply choose $C=B^{-1}$, then you get $$B^{-1}\vec{y} = B^{-1}B\vec{z} = \vec{z}$$ since $B^{-1}B = I$, where $I$ is the identity matrix.

Now we can plug this last result into the first equation: \begin{equation} \vec{x} = A\vec{z} = AB^{-1}\vec{y} \end{equation}

In your case we have $$A=\frac{L}{2}\left[\begin{array}{l}1&n\\-1&n\end{array}\right]$$ $$B=\frac{L}{2}\left[\begin{array}{l}n&1\\-n&1\end{array}\right]$$ It turns out that for $n\neq 0$ B is invertible and its inverse is $$B^{-1}=\frac{2}{L}\frac{1}{2n}\left[\begin{array}{l}1&-1\\n&n\end{array}\right]$$ Therefore for $\vec{x} = \left[\begin{array}{c}Lz\\y\end{array}\right]$, $\vec{y} = \left[\begin{array}{c}Lk\\m\end{array}\right]$ and $\vec{z} = \left[\begin{array}{c}P\\O\end{array}\right]$ As seen in the equation above we have $$\vec{x} = A\vec{z} = AB^{-1}\vec{y}$$ which becomes

$$\left[\begin{array}{c}Lz\\y\end{array}\right] = AB^{-1}\left[\begin{array}{c}Lk\\m\end{array}\right] = \frac{L}{2}\left[\begin{array}{l}1&n\\-1&n\end{array}\right]\frac{2}{L}\frac{1}{2n}\left[\begin{array}{l}1&-1\\n&n\end{array}\right]\left[\begin{array}{c}Lk\\m\end{array}\right] = \frac{1}{2n}\left[\begin{array}{l}1&n\\-1&n\end{array}\right]\left[\begin{array}{l}1&-1\\n&n\end{array}\right]\left[\begin{array}{c}Lk\\m\end{array}\right] $$

I hope this explanation was clear enough :-).

Answer 2

Guide:

From the second equation, we have

$$\begin{bmatrix} P \\ O \end{bmatrix}=\frac2L \begin{bmatrix} n & 1 \\ - n & 1\end{bmatrix}^{-1}\begin{bmatrix} Lk \\ m\end{bmatrix}$$

Substitute this into the first equation and you will get the desired identity.