Bilişim Matematiği - 7. Ders

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

Bir Kare Matrisin Tersi

Temel Formül

$$\begin{array}{c}\boxed{A\times A^{-1}=I}\end{array}$$

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Örnek 1:

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

Çözüm Adımları:

$$\begin{array}{l}\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \\ \to 2a+3c=12b+3d=0\\ \to a+2c=0b+2d=1\\ \\ \to (2a+3c)-(2a+4c)=1⟹-c=1\to c=-1\\ \to (2b+3d)-(2b+4d)=-2⟹-d=-2\to d=2\\ \\ \to a+2(-1)=0⟹a=2\\ \to b+2(2)=1⟹b=-3\\ \\ \to A^{-1}=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]\end{array}$$

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Örnek 2:

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

Çözüm Adımları:

$$\begin{array}{l}\left[\begin{array}{cc}-1& 3\\ 2& -4\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ \\ \to -a+3c=1-b+3d=0\\ \to 2a-4c=02b-4d=1\\ \\ \to c=1,a=2\\ \to 2d=1⟹d=\frac{1}{2}\\ \to -b+3(\frac{1}{2})=0⟹b=\frac{3}{2}\end{array}$$ $$A^{-1}=\left[\begin{array}{cc}2& \frac{3}{2}\\ 1& \frac{1}{2}\end{array}\right]$$

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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 3:

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

Çözüm Adımları:

$$\begin{array}{rl}& \left[\begin{array}{ccc}11& -1& 1\\ 5& 2& -1\\ 3& 1& -2\\ 11& -1& 1\\ 5& 2& -1\end{array}\right]⟹\begin{array}{r}|A|=(-44+5+3)-(6-11+10)\\ |A|=-36-5=-41\end{array}\end{array}$$

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Örnek 4:

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

Çözüm Adımları:

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

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Doğrusal Denklem Sistemlerinin Determinantlar Yardımıyla Çözümü (Cramer Yöntemi)

Örnek 5: Aşağıdaki denklemin çözüm kümesini Cramer yöntemiyle bulunuz.

$$\begin{array}{c}\text{3x + y = 11}\\ \text{x + y = 5}\end{array}$$

Çözüm Adımları:

$$\begin{array}{rl}& \Delta =\left|\begin{array}{cc}3& 1\\ 1& 1\end{array}\right|=3-1=2\\ & \\ & {\Delta }_x=\left|\begin{array}{cc}11& 1\\ 5& 1\end{array}\right|=6\to x=\frac{{\Delta }_x}{\Delta }=\frac{6}{2}=3\\ & \\ & {\Delta }_y=\left|\begin{array}{cc}3& 11\\ 1& 5\end{array}\right|=4\to y=\frac{{\Delta }_y}{\Delta }=2\\ & \end{array}$$ $$\text{C¸.K}=\{(3,2)\}$$

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Örnek 6: Aşağıdaki denklem sisteminin çözüm kümesini Cramer yöntemiyle bulunuz.

$$\begin{array}{r}2x-y+z=11\\ x+2y-z=5\\ -x+y-2z=3\end{array}$$

Çözüm Adımları:

$$\begin{array}{rl}\Delta =& \left|\begin{array}{rrr}2& -1& 1\\ 1& 2& -1\\ -1& 1& -2\end{array}\right|\\ & \begin{array}{rrr}2& -1& 1\\ 1& 2& -1\end{array}\end{array}\begin{array}{l}\to (-8+1-1)-(-2-2+2)\\ =-8+2=-6\end{array}$$ $$\begin{array}{rl}{\Delta }_x=& \left|\begin{array}{rrr}11& -1& 1\\ 5& 2& -1\\ 3& 1& -2\end{array}\right|\\ & \begin{array}{rrr}11& -1& 1\\ 5& 2& -1\end{array}\end{array}\begin{array}{l}\to (-44+5+3)-(6-11+10)\\ =-36-5=-41\\ x=\frac{{\Delta }_x}{\Delta }=\frac{-41}{6}\end{array}$$ $$\begin{array}{rl}{\Delta }_y=& \left|\begin{array}{rrr}2& 11& 1\\ 1& 5& -1\\ -1& 3& -2\end{array}\right|\\ & \begin{array}{rrr}2& 11& 1\\ 1& 5& -1\end{array}\end{array}\begin{array}{l}\to (-20+11+3)-(-5-6-22)\\ =(-6)-(-33)=27\\ y=\frac{{\Delta }_y}{\Delta }=\frac{27}{-6}=-\frac{9}{2}\end{array}$$ $$\begin{array}{rl}{\Delta }_z=& \left|\begin{array}{rrr}2& -1& 11\\ 1& 2& 5\\ -1& 1& 3\end{array}\right|\\ & \begin{array}{rrr}2& -1& 11\\ 1& 2& 5\end{array}\end{array}\begin{array}{l}\to (12+5+11)-(-22+10-3)\\ =28-(-15)=43\\ z=\frac{{\Delta }_z}{\Delta }=\frac{43}{-6}=-\frac{43}{6}\end{array}$$
$$\text{C¸.K}=\left\{\left(\frac{41}{6},-\frac{9}{2},-\frac{43}{6}\right)\right\}$$