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Sagot :
[tex]\begin{gathered} 4x-12=-1 \\ 6x+4y=4 \end{gathered}[/tex]
We want to solve this system using a matrix.
The first step is write a matrix whose each line will correspond to a equation and each term will correspond to a coeficient. The terms of the firs colum will correspond to the coeficients of x, the terms of the second colum will correspond to the coeficients of y and the terms of the third colum will correspond to the independent term in the right side of the equation:
[tex]\begin{bmatrix}{4} & {-12} & {-1} \\ {6} & {4} & {4} \\ {\square} & {\square} & {\square}\end{bmatrix}[/tex]Now, we must conduct operations to escalonate the terms corresponding to the coefficients multiplying x and y
First, we multiply the second line by 3:
[tex]\begin{bmatrix}{4} & {-12} & {-1} \\ {18} & {12} & {12} \\ {\square} & {\square} & {\square}\end{bmatrix}[/tex]Then, we add line 2 to line 1:
[tex]\begin{bmatrix}{22} & {0} & {11} \\ {18} & {12} & {12} \\ {\square} & {\square} & {\square}\end{bmatrix}[/tex]Now, we divide line 1 by 22
[tex]\begin{bmatrix}{1} & {0} & {\frac{1}{2}} \\ {18} & {12} & {12} \\ {\square} & {\square} & {\square}\end{bmatrix}[/tex]Then, we subtract 18 times line 1 from line 2:
[tex]\begin{gathered} \begin{bmatrix}{1} & {0} & {\frac{1}{2}} \\ {18-18} & {12} & {12-\frac{18}{2}} \\ {\square} & {\square} & {\square}\end{bmatrix} \\ \begin{bmatrix}{1} & 0 & {\frac{1}{2}} \\ {0} & {12} & {3} \\ {\square} & {\square} & {\square}\end{bmatrix} \end{gathered}[/tex]Now, we divide line 2 by 12:
[tex]\begin{bmatrix}{1} & {0} & {\frac{1}{2}} \\ {0} & {1} & {\frac{1}{4}} \\ {\square} & {\square} & {\square}\end{bmatrix}[/tex]Finally, we can rewrite these terms in the form of equations and obtain the solution:
[tex]\begin{gathered} x=\frac{1}{2} \\ y=\frac{1}{4} \end{gathered}[/tex]
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