We are given the equations:
x - 3y = 2
3x - 4y = 0
writing the system as matrices
![\left[\begin{array}{ccc}1&-3\\3&-4\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/xm43ktt34q699rdfhwzht5jbpbkbvzy7pq.png)
=
![\left[\begin{array}{ccc}2\\0\\\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/aahdbkwtyt697x6twi83443nx8d6qnijzm.png)
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
![\left[\begin{array}{ccc}w&x\\y&z\end{array}\right]^(-1) = (1)/(wz-xy)\left[\begin{array}{ccc}z&-x\\-y&w\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/1tea73b58kfzgvrsraofupxul46xzpjc16.png)
using this formula to find the inverse of matrix A:
![A^(-1) = (1)/((1*-4)-(-3*3))\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/hh72z8y3t6oqkimdfu9087plifgnyz796z.png)
![A^(-1) = (1)/(5)\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/jsnomp8jr0ppao05f8r3dqtw9smj8d0yim.png)
Matrix X:
We know that:
X = A⁻¹B
![X = (1)/(5)\left[\begin{array}{ccc}-4&3\\-3&1\end{array}\right] * \left[\begin{array}{ccc}2\\0\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/45ta76vy6y5zhc9plgmluzr9hppazy316b.png)
![X = (1)/(5)\left[\begin{array}{ccc}-8\\-6\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/im5q01dhfhjy73rbw925ms3o58nsqmi24g.png)
since matrix X is just a matrix with the variables
![\left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}(-8)/(5)\\(-6)/(5)\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/college/7wbdb1bv0petaszlrlk1b9ssxlqdz7bhb4.png)
x = -8/5
y = -6/5