Answer:
[tex]X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}[/tex]
Step-by-step explanation:
The question we have at hand is, in other words,
[tex]\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}[/tex] - where we have to solve for the value of [tex]X[/tex]
If we have to isolate [tex]X[/tex] here, then we would have to take the inverse of the following matrix ...
[tex]\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}[/tex] ... so that it should be as follows ... [tex]\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}[/tex]
Therefore, we can conclude that the equation as to solve for " [tex]X[/tex] " will be the following,
[tex]X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}[/tex] - First find the 2 x 2 matrix inverse of the first portion,
[tex]\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}[/tex] = [tex]\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}[/tex]= [tex]\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}[/tex] = [tex]\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}[/tex]
At this point we have to multiply the rows of the first matrix by the rows of the second matrix,
[tex]X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}[/tex] ,
[tex]X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}[/tex] - Simplifying this, we should get ...
[tex]\begin{bmatrix}1&3\\ 2&4\end{bmatrix}[/tex] ... which is our solution.
