२ अज्ञात राशि वाले २ रैखिक समीकरणों का निकाय संपादित करें
माना दो अज्ञात राशि से युक्त दो रैखिक समीकरण ये हैं:
{
a
x
+
b
y
=
e
c
x
+
d
y
=
f
{\displaystyle {\begin{cases}a{\color {blue}x}+b{\color {blue}y}={\color {red}e}\\c{\color {blue}x}+d{\color {blue}y}={\color {red}f}\end{cases}}}
इनका मैट्रिक्स निरूपण यह है:
[
a
b
c
d
]
[
x
y
]
=
[
e
f
]
{\displaystyle {\begin{bmatrix}a&b\\c&d\end{bmatrix}}{\begin{bmatrix}{\color {blue}x}\\{\color {blue}y}\end{bmatrix}}={\begin{bmatrix}{\color {red}e}\\{\color {red}f}\end{bmatrix}}}
क्रैमर का नियम लगाकर
x
{\displaystyle x}
तथा
y
{\displaystyle y}
का मान यह निकलता है:
x
=
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e
b
f
d
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|
a
b
c
d
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=
e
d
−
b
f
a
d
−
b
c
;
y
=
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a
e
c
f
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a
b
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=
a
f
−
e
c
a
d
−
b
c
{\displaystyle x={\frac {\begin{vmatrix}\color {red}{e}&b\\\color {red}{f}&d\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}}={\frac {{\color {red}e}d-b{\color {red}f}}{ad-bc}};\quad y={\frac {\begin{vmatrix}a&\color {red}{e}\\c&\color {red}{f}\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}}={\frac {a{\color {red}f}-{\color {red}e}c}{ad-bc}}}
3
x
+
1
y
=
9
{\displaystyle 3x+1y=9\,}
2
x
+
3
y
=
13
{\displaystyle 2x+3y=13\,}
इनको मैट्रिक्स रूप में लिखने पर:
[
3
1
2
3
]
[
x
y
]
=
[
9
13
]
{\displaystyle {\begin{bmatrix}3&1\\2&3\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}9\\13\end{bmatrix}}}
क्रैमर नियम से x और y का मान यह है:
x
=
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9
1
13
3
|
|
3
1
2
3
|
=
9
∗
3
−
1
∗
13
3
∗
3
−
1
∗
2
=
2
{\displaystyle x={\frac {\begin{vmatrix}9&1\\13&3\end{vmatrix}}{\begin{vmatrix}3&1\\2&3\end{vmatrix}}}={9*3-1*13 \over 3*3-1*2}=2}
5
y
=
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3
9
2
13
|
|
3
1
2
3
|
=
3
∗
13
−
9
∗
2
3
∗
3
−
1
∗
2
=
3
{\displaystyle y={\frac {\begin{vmatrix}3&9\\2&13\end{vmatrix}}{\begin{vmatrix}3&1\\2&3\end{vmatrix}}}={3*13-9*2 \over 3*3-1*2}=3}
8
माना मैट्रिक्स रूप में निरूपित 3x3 रैखिक समीकरण निकाय यह है:
{
a
x
+
b
y
+
c
z
=
j
d
x
+
e
y
+
f
z
=
k
g
x
+
h
y
+
i
z
=
l
{\displaystyle {\begin{cases}a{\color {blue}x}+b{\color {blue}y}+c{\color {blue}z}={\color {black}j}\\d{\color {blue}x}+e{\color {blue}y}+f{\color {blue}z}={\color {black}k}\\g{\color {blue}x}+h{\color {blue}y}+i{\color {blue}z}={\color {black}l}\end{cases}}}
इसका हल यह है:
[
a
b
c
d
e
f
g
h
i
]
[
x
y
z
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=
[
j
k
l
]
{\displaystyle {\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}}{\begin{bmatrix}{\color {blue}x}\\{\color {blue}y}\\{\color {blue}z}\end{bmatrix}}={\begin{bmatrix}{\color {red}j}\\{\color {red}k}\\{\color {red}l}\end{bmatrix}}}
x
{\displaystyle x}
,
y
{\displaystyle y}
,
z
{\displaystyle z}
pueden ser encontradas como sigue:
x
=
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j
b
c
k
e
f
l
h
i
|
|
a
b
c
d
e
f
g
h
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;
y
=
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a
j
c
d
k
f
g
l
i
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a
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c
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e
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g
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,
z
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d
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k
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l
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a
b
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g
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{\displaystyle x={\frac {\begin{vmatrix}{\color {red}j}&b&c\\{\color {red}k}&e&f\\{\color {red}l}&h&i\end{vmatrix}}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}};\quad y={\frac {\begin{vmatrix}a&{\color {red}j}&c\\d&{\color {red}k}&f\\g&{\color {red}l}&i\end{vmatrix}}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}},\quad z={\frac {\begin{vmatrix}a&b&{\color {red}j}\\d&e&{\color {red}k}\\g&h&{\color {red}l}\end{vmatrix}}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}}}
3
x
+
2
y
+
1
z
=
1
{\displaystyle 3x+2y+1z=1\,}
2
x
+
0
y
+
1
z
=
2
{\displaystyle 2x+0y+1z=2\,}
−
1
x
+
1
y
+
2
z
=
4
{\displaystyle -1x+1y+2z=4\,}
मैट्रिक्स रूप में लिखने पर:
[
3
2
1
2
0
1
−
1
1
2
]
[
x
y
z
]
=
[
1
2
4
]
{\displaystyle {\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}}}
x
,
y
y
z
{\displaystyle x,y{\text{ y }}z}
के मान ये होंगे:
x
=
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1
2
1
2
0
1
4
1
2
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|
3
2
1
2
0
1
−
1
1
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;
y
=
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3
1
1
2
2
1
−
1
4
2
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|
3
2
1
2
0
1
−
1
1
2
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;
z
=
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3
2
1
2
0
2
−
1
1
4
|
|
3
2
1
2
0
1
−
1
1
2
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{\displaystyle x={\frac {\begin{vmatrix}1&2&1\\2&0&1\\4&1&2\end{vmatrix}}{\begin{vmatrix}3&2&1\\2&0&1\\-1&1&2\end{vmatrix}}};\quad y={\frac {\begin{vmatrix}3&1&1\\2&2&1\\-1&4&2\end{vmatrix}}{\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}}}}