19 lines
984 B
Markdown
19 lines
984 B
Markdown
#s/maths/algèbre
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---
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Soit le système :
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$(S) :\left\{ \begin{gathered}ax+by = c\\ a'x + b'y = c' \end{gathered}\right.$
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$$\begin{align*}
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(S) & \iff \begin{pmatrix} a&b\\a'&b' \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} c \\ c'\end{pmatrix}\\
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&\iff \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}a&b\\a'&b'\end{pmatrix}^{-1}\begin{pmatrix}c\\c'\end{pmatrix}\\
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&\iff \begin{pmatrix}x\\y\end{pmatrix} = \frac{1}{\left|\small \begin{matrix}a&b\\a'&b'\end{matrix} \right|} \begin{pmatrix}cb' - c'b\\c'a-ca'\end{pmatrix}\\
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&\iff \left\{\begin{gathered}
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x = \frac{1}{\left|\small \begin{matrix}a &b\\a'&b'\end{matrix}\right|} \left| \begin{matrix}c & b\\ c' & b'\end{matrix} \right| \\
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y = \frac{1}{\left|\small \begin{matrix}a &b\\a'&b'\end{matrix}\right|} \left| \begin{matrix}a&c\\a'&c'\end{matrix} \right|
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\end{gathered}\right.
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&\end{align*}$$
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Donc, si $\begin{pmatrix} a&b\\a'&b' \end{pmatrix}^{-1}$ existe, il y a une unique solution au système $(S)$
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