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Modifications apportées au TD2 de mathématiques générales

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Feror 2026-04-02 11:06:07 +02:00
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MathématiquesGénérales/CI-SST81E6_TD2.pdf Normal file
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MathématiquesGénérales/TD1.md
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@ -174,7 +174,7 @@ On a donc $k_1=-\frac{-x+1}{2}$ et $k_2=(x+1)(-\frac{-x+1}{2})$.
$q_p=-1+(x^2+1)(\frac{-x+1}{2})=+\frac{1}{2}(-x^3+x^2-x-1)$ $q_p=-1+(x^2+1)(\frac{-x+1}{2})=+\frac{1}{2}(-x^3+x^2-x-1)$
$q_h=(x^2+1)(x-1)k\,\forall k\in\mathbb{R}[x]$. $q_h=(x^2+1)(x-1)k\,\forall k\in\mathbb{R}[x]$.
$q=q_p+q_h=-\frac{1}{2}x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{2}+k(x^3-x^2+x-1)$ $q=q_p+q_h=-\frac{1}{2}x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{2}+k(x^3-x^2+x-1)$
$\cases{q\equiv\frac{1}{2}(-x^3+x^2-x-1)[x^3-x^2+x-1] \\ q\equiv1[x^2-x+1]}$ $\cases{q\equiv\frac{1}{2}(-x^3+x^2-x-1)[x^3-x^2+x-1] \\ q\equiv1[x^2-x+1]}$
$\exists k_3\in\mathbb{R}[x]$ tel que $q=1+k_3(x^2-x+1)$. $\exists k_3\in\mathbb{R}[x]$ tel que $q=1+k_3(x^2-x+1)$.

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MathématiquesGénérales/TD2.md
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@ -1,10 +1,12 @@
![[CI-SST81E6_TD2.pdf]]
# Exercice 1: # Exercice 1:
$A=\pmatrix{-1&0&1\\1&0&-1\\0&0&0}$ $A=\pmatrix{-1&0&1\\1&0&-1\\0&0&0}$
## 1) ## 1)
$X_A(x)=\det{(x\times I - A)}=\matrix{+\\-\\+}\begin{vmatrix}x+1&0&-1\\-1&x&1\\0&0&x\end{vmatrix}=x^3+x^x=x^2(x+1)$ $X_A(x)=\det{(x\times I - A)}=\matrix{+\\-\\+}\begin{vmatrix}x+1&0&-1\\-1&x&1\\0&0&x\end{vmatrix}=x^3+x^x=x^2(x+1)$
## 2) ## 2)
$A\times (A+I)$ On veut calculer $A\times (A+I)$
$A+I=\pmatrix{0&0&1\\1&1&1\\0&0&1}\pmatrix{-1&0&1\\1&0&-1\\0&0&0}=\pmatrix{0&0&0\\0&0&0\\0&0&0}$ $A+I=\pmatrix{0&0&1\\1&1&-1\\0&0&1}$
$A\times (A+I)=\pmatrix{0&0&1\\1&1&-1\\0&0&1}\pmatrix{-1&0&1\\1&0&-1\\0&0&0}=\pmatrix{0&0&0\\0&0&0\\0&0&0}$
$p:x\rightarrow x(x+1)$. $p:x\rightarrow x(x+1)$.
$p$ est le polynome minimal de $A$. $p$ est le polynome minimal de $A$.