We are given the equations:
x - 3y = 2
3x - 4y = 0
writing the system as matrices
[tex]\left[\begin{array}{ccc}1&-3\\3&-4\end{array}\right][/tex][tex]\left[\begin{array}{ccc}x\\y\end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}2\\0\\\end{array}\right][/tex]
which is in the form:
AX = B
solving for X(the matrix holding the variables), we get:
X = (A⁻¹)B
Finding A⁻¹:
now, to do this, we need to find the inverse of A
[tex]\left[\begin{array}{ccc}w&x\\y&z\end{array}\right]^{-1} = \frac{1}{wz-xy}\left[\begin{array}{ccc}z&-x\\-y&w\end{array}\right][/tex]
using this formula to find the inverse of matrix A:
[tex]A^{-1} = \frac{1}{(1*-4)-(-3*3)}\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right][/tex]
[tex]A^{-1} = \frac{1}{5}\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right][/tex]
Matrix X:
We know that:
X = A⁻¹B
[tex]X = \frac{1}{5}\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\end{array}\right][/tex]
[tex]X = \frac{1}{5}\left[\begin{array}{ccc}-8\\-6\end{array}\right][/tex]
since matrix X is just a matrix with the variables
[tex]\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}\frac{-8}{5}\\\frac{-6}{5}\end{array}\right][/tex]
x = -8/5
y = -6/5
