लीनियर एलजेब्रा उदाहरण
S([abc])=[a-2b-c3a-b+2ca+b+2c]S⎛⎜⎝⎡⎢⎣abc⎤⎥⎦⎞⎟⎠=⎡⎢⎣a−2b−c3a−b+2ca+b+2c⎤⎥⎦
चरण 1
परिवर्तन का कर्नेल एक सदिश है जो परिवर्तन को शून्य सदिश (रूपांतरण की पूर्व-छवि) के बराबर बनाता है.
[a-2b-c3a-b+2ca+b+2c]=0⎡⎢⎣a−2b−c3a−b+2ca+b+2c⎤⎥⎦=0
चरण 2
सदिश समीकरण से समीकरणों की एक प्रणाली बनाएंँ.
a-2b-c=0a−2b−c=0
3a-b+2c=03a−b+2c=0
a+b+2c=0a+b+2c=0
चरण 3
Write the system as a matrix.
[1-2-103-1201120]⎡⎢
⎢⎣1−2−103−1201120⎤⎥
⎥⎦
चरण 4
चरण 4.1
Perform the row operation R2=R2-3R1R2=R2−3R1 to make the entry at 2,12,1 a 00.
चरण 4.1.1
Perform the row operation R2=R2-3R1R2=R2−3R1 to make the entry at 2,12,1 a 00.
[1-2-103-3⋅1-1-3⋅-22-3⋅-10-3⋅01120]⎡⎢
⎢⎣1−2−103−3⋅1−1−3⋅−22−3⋅−10−3⋅01120⎤⎥
⎥⎦
चरण 4.1.2
R2R2 को सरल करें.
[1-2-1005501120]⎡⎢
⎢⎣1−2−1005501120⎤⎥
⎥⎦
[1-2-1005501120]⎡⎢
⎢⎣1−2−1005501120⎤⎥
⎥⎦
चरण 4.2
Perform the row operation R3=R3-R1R3=R3−R1 to make the entry at 3,13,1 a 00.
चरण 4.2.1
Perform the row operation R3=R3-R1R3=R3−R1 to make the entry at 3,13,1 a 00.
[1-2-1005501-11+22+10-0]⎡⎢
⎢⎣1−2−1005501−11+22+10−0⎤⎥
⎥⎦
चरण 4.2.2
R3R3 को सरल करें.
[1-2-1005500330]⎡⎢
⎢⎣1−2−1005500330⎤⎥
⎥⎦
[1-2-1005500330]⎡⎢
⎢⎣1−2−1005500330⎤⎥
⎥⎦
चरण 4.3
Multiply each element of R2R2 by 1515 to make the entry at 2,22,2 a 11.
चरण 4.3.1
Multiply each element of R2R2 by 1515 to make the entry at 2,22,2 a 11.
[1-2-10055555050330]⎡⎢
⎢⎣1−2−10055555050330⎤⎥
⎥⎦
चरण 4.3.2
R2R2 को सरल करें.
[1-2-1001100330]⎡⎢
⎢⎣1−2−1001100330⎤⎥
⎥⎦
[1-2-1001100330]⎡⎢
⎢⎣1−2−1001100330⎤⎥
⎥⎦
चरण 4.4
Perform the row operation R3=R3-3R2R3=R3−3R2 to make the entry at 3,23,2 a 00.
चरण 4.4.1
Perform the row operation R3=R3-3R2R3=R3−3R2 to make the entry at 3,23,2 a 0.
[1-2-1001100-3⋅03-3⋅13-3⋅10-3⋅0]
चरण 4.4.2
R3 को सरल करें.
[1-2-1001100000]
[1-2-1001100000]
चरण 4.5
Perform the row operation R1=R1+2R2 to make the entry at 1,2 a 0.
चरण 4.5.1
Perform the row operation R1=R1+2R2 to make the entry at 1,2 a 0.
[1+2⋅0-2+2⋅1-1+2⋅10+2⋅001100000]
चरण 4.5.2
R1 को सरल करें.
[101001100000]
[101001100000]
[101001100000]
चरण 5
Use the result matrix to declare the final solution to the system of equations.
a+c=0
b+c=0
0=0
चरण 6
Write a solution vector by solving in terms of the free variables in each row.
[abc]=[-c-cc]
चरण 7
Write the solution as a linear combination of vectors.
[abc]=c[-1-11]
चरण 8
Write as a solution set.
{c[-1-11]|c∈R}
चरण 9
The solution is the set of vectors created from the free variables of the system.
{[-1-11]}
चरण 10
S का अष्टि उप-स्थान {[-1-11]} है.
K(S)={[-1-11]}
