Bilişim Matematiği - 7. Hafta

Ünite 4: Lineer Cebir (Matrisler ve Determinantlar) | 07.11.2025

Bir Kare Matrisin Tersi

$\boxed{A\times A^{-1}=I}$

Örnek 1

$A=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]⟹A^{-1}=?$

$\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\to 2a+3c=12b+3d=0$

$\to a+2c=0b+2d=1$

\begin{array} & &&&&&&&&&&&&&&&&& \\ \hline \end{array}

$\to 2a+3c-(2a+4c)=1⟹-c=1\to$

$c=-1$

$\to 2b+3d-(2b+4d)=-2⟹-d=-2$

$d=2$

$\to a-2=0⟹a=2\wedge b+4=1⟹b=-3$

$\to A^{-1}=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$

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Örnek 2

$\left[\begin{array}{cc}-1& 3\\ 2& -4\end{array}\right]⟹A^{-1}=?$

$\left[\begin{array}{cc}-1& 3\\ 2& -4\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\to -a+3c=1-b+3d=0$

$\to 2a-4c=02b-4d=1$

\begin{array} & &&&&&&&&&&&&&&&&& \\ \hline \end{array}

$c=1,a=2$

$2d=1⟹d=\frac{1}{2}$

$b-2d=1⟹b=\frac{3}{2}$

$A^{-1}=\left[\begin{array}{cc}2& \frac{3}{2}\\ 1& \frac{1}{2}\end{array}\right]$

Determinantlar

• $n\in {\mathbb{N}}^{+}$ için, $n$. mertebeden bütün kare matrislerden reel sayılara tanımlı $|A|$ ise, bu tür ifade edilen fonksiyonlara determinant fonksiyonu denir.

Bir Kare Matrisin Determinantının Alınması (Saruss Kuralı)

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Örnek 1

$A=\left[\begin{array}{ccc}11& -1& 1\\ 5& 2& -1\\ 3& 1& -2\end{array}\right]⟹|A|=?$

$\left[\begin{array}{ccc}11& -1& 1\\ 5& 2& -1\\ 3& 1& -2\\ 11& -1& 1\\ 5& 2& -1\end{array}\right]⟹$

$\to |A|=(-44+5+3)-(6-11+10)$

$\to |A|=-36-5=-41$

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Örnek 2

$A=\left[\begin{array}{ccc}4& 0& 5\\ 3& 6& -1\\ 2& 0& -3\end{array}\right]⟹|A|=?$

$\left[\begin{array}{ccc}4& 0& 5\\ 3& 6& -1\\ 2& 0& -3\\ 4& 0& 5\\ 3& 6& -1\end{array}\right]⟹|A|=(-72+0+0)-(60+0+0)=-132$

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Doğrusal Denklem Sistemlerinin Determinantlar Yardımıyla Çözümü (Cramer Yöntemi)

Örnek 1

Aşağıdaki denklemin çözüm kümesini Cramer yöntemiyle bulunuz. $\begin{array}{c}\text{3x + y = 11}\\ \text{x}\text{+y = 5}\end{array}$ $\Delta =\left|\begin{array}{cc}3& 1\\ 1& 1\end{array}\right|=3-1=2$

${\Delta }_x=\left|\begin{array}{cc}11& 1\\ 5& 1\end{array}\right|=6\to x=\frac{{\Delta }_x}{\Delta }=\frac{6}{2}=3$

${\Delta }_y=\left|\begin{array}{cc}3& 1\\ 11& 5\end{array}\right|=4\to y=\frac{{\Delta }_y}{\Delta }=2$

$\stackrel{\backslash c}{C}.K=\{(3,2)\}$

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Örnek 2

Aşağıdaki denklem sisteminin çözüm kümesini Cramer yönetimiyle bulunuz.