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Sagot :
We have the following 2x2 system of equations:
[tex]\begin{pmatrix}4 & -12 \\ 6 & 4\end{pmatrix}\cdot\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}-1 \\ 4\end{pmatrix}[/tex]and we need the vector solution
[tex]\begin{pmatrix}x \\ y\end{pmatrix}[/tex]Then, we need to find the inverse matrix of the 2x2 matrix on the left hand side. For any 2x2 matrix A,
[tex]A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}[/tex]the inverse is given as.
[tex]A^{-1}=\frac{1}{\det A}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}[/tex]where detA denotes the determinant. Therefore, by means of the inverse matrix, the general solution for any 2x2 matrix will be
[tex]\begin{pmatrix}x \\ y\end{pmatrix}=A^{-1}\begin{pmatrix}z \\ w\end{pmatrix}[/tex]for any vector with entries z and w.
In our case, the determinat is
[tex]\begin{gathered} \det \begin{pmatrix}4 & -12 \\ 6 & 4\end{pmatrix}=4\times4-(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}[/tex]Therefore, the solution of our system will be
[tex]\begin{gathered} \begin{pmatrix}x \\ y\end{pmatrix}=A^{-1}\begin{pmatrix}-1 \\ 4\end{pmatrix} \\ \text{with } \\ A^{-1}=\frac{1}{88}\begin{pmatrix}4 & 12 \\ -6 & 4\end{pmatrix} \end{gathered}[/tex]Explicitly,
[tex]\begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{88}\begin{pmatrix}4 & 12 \\ -6 & 4\end{pmatrix}\cdot\begin{pmatrix}-1 \\ 4\end{pmatrix}[/tex]Now, lets make the product of the right hand side. It yields,
[tex]\begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{88}\begin{pmatrix}4(-1)+(12)(4) \\ (-6)(-1)+(4)(4)\end{pmatrix}[/tex]which gives
[tex]\begin{gathered} \begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{88}\begin{pmatrix}-4+48 \\ 6+16\end{pmatrix} \\ \begin{pmatrix}x \\ y\end{pmatrix}=\frac{1}{88}\begin{pmatrix}44 \\ 22\end{pmatrix} \end{gathered}[/tex]since 22x4=48 and 44x2=88, we have
[tex]\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}\frac{1}{2} \\ \frac{1}{4}\end{pmatrix}[/tex]Therefore, the solution of the system is
[tex]\begin{gathered} x=\frac{1}{2}=0.5 \\ \text{and} \\ y=\frac{1}{4}=0.25 \end{gathered}[/tex]
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