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== क्रैमर का नियम ==
{{मुख्य|क्रैमर-नियम}}
रैखिक समीकरणों के निकाय का हल निकलने के लिए सन् १७५० में क्रैमर ने एक प्रत्यक्ष विधि (direct method) बताया। यह गुणाण्क मैट्रिक्स के व्युत्क्रमण (इन्वर्सन) पर आधारित है।
=== उदाहरण (२)===
:<math>3x+2y+1z = 1\begin{matrix},</math>
:<math>2x+0y+1z = 2\,</math>
\color{blue}{82}\,\color{black}x_1+\color{blue}{45}\,\color{black}x_2+\color{blue}{9}\,\color{black}x_3=\color{OliveGreen}{1}\\
:<math>-1x+1y+2z = 4\,</math>
\color{blue}{27}\,\color{black}x_1+\color{blue}{16}\,\color{black}x_2+\color{blue}{3}\,\color{black}x_3=\color{OliveGreen}{1}\\
\color{blue}{9}\,\color{black}x_1+\color{blue}{5}\,\color{black}x_2+\color{blue}{1}\,\color{black}x_3=\color{OliveGreen}{0}\\
\end{matrix}</math>
:<math>x_1 = \frac{\det(A_1)}{\det(A)} =
\frac{\begin{vmatrix}\color{OliveGreen}{1}
&\color{blue}{45}
&\color{blue}{9}
\\ \color{OliveGreen}{1}
&\color{blue}{16}
&\color{blue}{3}
\\ \color{OliveGreen}{0}
&\color{blue}{5}
&\color{blue}{1}
\end{vmatrix}}
{\begin{vmatrix}\color{blue}{82}
&\color{blue}{45}
&\color{blue}{9}
\\ \color{blue}{27}
&\color{blue}{16}
&\color{blue}{3}
\\ \color{blue}{9}
&\color{blue}{5}
&\color{blue}{1}
\end{vmatrix}}
= \frac{1}{1} = 1\qquad</math>
:<math>x_2 = \frac{\det(A_2)}{\det(A)} =
मैट्रिक्स रूप में लिखने पर:
\frac{\begin{vmatrix}\color{blue}{82}
&\color{OliveGreen}{1}
&\color{blue}{9}
\\ \color{blue}{27}
&\color{OliveGreen}{1}
&\color{blue}{3}
\\ \color{blue}{9}
&\color{OliveGreen}{0}
&\color{blue}{1}
\end{vmatrix}}
{\begin{vmatrix}\color{blue}{82}
&\color{blue}{45}
&\color{blue}{9}
\\ \color{blue}{27}
&\color{blue}{16}
&\color{blue}{3}
\\ \color{blue}{9}
&\color{blue}{5}
&\color{blue}{1}
\end{vmatrix}}
= \frac{1}{1} = 1\qquad</math>
<math>
:<math>x_3 = \frac{\det(A_3)}{\det(A)} =
\begin{bmatrix}
3 & 2 & 1 \\
2 & 0 & 1 \\
-1 & 1 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
1\\
2\\
4
\end{bmatrix}
</math>
<math>x, y \text{ y } z</math> के मान ये होंगे:
\frac{\begin{vmatrix}\color{blue}{82}
:<math> x= \frac {
&\color{blue}{45}
\begin{vmatrix}
&\color{Olive Green}{1}
1 & 2 & 1\\
\\ \color{blue}{27}
2 & 0 & 1\\
&\color{blue}{16}
4 & 1 & 2
&\color{OliveGreen}{1}
\end{vmatrix}
\\ \color{blue}{9}
}{
&\color{blue}{5}
\begin{vmatrix}
&\color{OliveGreen}{0}
3 & 2 & 1\\
\end{vmatrix}}
2 & 0 & 1\\
{\begin{vmatrix}\color{blue}{82}
-1 & 1 & 2
&\color{blue}{45}
\end{vmatrix}
&\color{blue}{9}
} ; \quad
\\ \color{blue}{27}
y= \frac {
&\color{blue}{16}
\begin{vmatrix}
&\color{blue}{3}
3 & 1 & 1\\
\\ \color{blue}{9}
2 & 2 & 1\\
&\color{blue}{5}
-1 & 4 & 2
&\color{blue}{1}
\end{vmatrix}}
}{
= \frac{-14}{1} = -14</math>
\begin{vmatrix}
3 & 2 & 1\\
2 & 0 & 1\\
-1 & 1 & 2
\end{vmatrix}
} ; \quad
z= \frac {
\begin{vmatrix}
3 & 2 & 1\\
2 & 0 & 2\\
-1 & 1 & 4
\end{vmatrix}
}{
\begin{vmatrix}
3 & 2 & 1\\
2 & 0 & 1\\
-1 & 1 & 2
\end{vmatrix}
}
</math>
== सन्दर्भ ==
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