We have the following 2x2 system of equations:
![\begin{pmatrix}4 & -12 \\ 6 & 4\end{pmatrix}\cdot\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}-1 \\ 4\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/jslwwk8sqqntcxyh2ub2ureg7npvz25s8l.png)
and we need the vector solution
![\begin{pmatrix}x \\ y\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/a0utqb3hfu2sp312azsfw2ml7xnp7ipw7t.png)
Then, we need to find the inverse matrix of the 2x2 matrix on the left hand side. For any 2x2 matrix A,
![A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/79e5aqw7a0pbutuxes6fzrrynsn0i2oemg.png)
the inverse is given as.
![A^(-1)=(1)/(\det A)\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/7gcvwa3fg2iihtbcixpa2cm37pnswm9cg6.png)
where detA denotes the determinant. Therefore, by means of the inverse matrix, the general solution for any 2x2 matrix will be
![\begin{pmatrix}x \\ y\end{pmatrix}=A^(-1)\begin{pmatrix}z \\ w\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/gbaxb907qzajnvog0apu939lxkf1k3qp3h.png)
for any vector with entries z and w.
In our case, the determinat is
![\begin{gathered} \det \begin{pmatrix}4 & -12 \\ 6 & 4\end{pmatrix}=4*4-(6)(-12) \\ \det \begin{pmatrix}4 & -12 \\ 6 & 4\end{pmatrix}=16+72 \\ \det \begin{pmatrix}4 & -12 \\ 6 & 4\end{pmatrix}=88 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2buaatvkzrwfqmgzik346vu8h2n12brs92.png)
Therefore, the solution of our system will be
![\begin{gathered} \begin{pmatrix}x \\ y\end{pmatrix}=A^(-1)\begin{pmatrix}-1 \\ 4\end{pmatrix} \\ \text{with } \\ A^(-1)=(1)/(88)\begin{pmatrix}4 & 12 \\ -6 & 4\end{pmatrix} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h2i025r630zm9k93m62lm5uyxwmqbox41n.png)
Explicitly,
![\begin{pmatrix}x \\ y\end{pmatrix}=(1)/(88)\begin{pmatrix}4 & 12 \\ -6 & 4\end{pmatrix}\cdot\begin{pmatrix}-1 \\ 4\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/pu34h8ly8cs8v0e4vy66x1bn4dtodt44k5.png)
Now, lets make the product of the right hand side. It yields,
![\begin{pmatrix}x \\ y\end{pmatrix}=(1)/(88)\begin{pmatrix}4(-1)+(12)(4) \\ (-6)(-1)+(4)(4)\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/ecjbccd0g6chvhgc2rma7hfcg7tzf3hrbb.png)
which gives
![\begin{gathered} \begin{pmatrix}x \\ y\end{pmatrix}=(1)/(88)\begin{pmatrix}-4+48 \\ 6+16\end{pmatrix} \\ \begin{pmatrix}x \\ y\end{pmatrix}=(1)/(88)\begin{pmatrix}44 \\ 22\end{pmatrix} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wsojthtnqeeof8mvgsbvvcnqmqn3qkpy7t.png)
since 22x4=48 and 44x2=88, we have
![\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}(1)/(2) \\ (1)/(4)\end{pmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/duaunx86dt88vwke3lzp2u8lbbig6oggqy.png)
Therefore, the solution of the system is
![\begin{gathered} x=(1)/(2)=0.5 \\ \text{and} \\ y=(1)/(4)=0.25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qryay72q33iumu4lrgg24jyvok1hhpfqi0.png)