Solve the following system of linear equations, by Gaussian elimination method :
4x+3y +6z=25 ; x+5y+7z=13;2x+9y+z=1
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Answer:
Let the matrix A be
⎣
⎢
⎢
⎡
4
1
2
3
5
9
6
7
1
⎦
⎥
⎥
⎤
and matrix B be
⎣
⎢
⎢
⎡
25
13
1
⎦
⎥
⎥
⎤
∣A∣
=0 , therefore the rank of A is 3
Now consider A∣B=
⎣
⎢
⎢
⎡
4
1
2
3
5
9
6
7
1
25
13
1
⎦
⎥
⎥
⎤
, the rank is 3
Therefore ρ(A)=ρ(A∣B)=3 , which implies that the system of equations is consistent.
ρ(A)=ρ(A∣B)= number of rows , therefore the sytem has unique solution.
We will solve it by determinant method.
The value of D=
∣
∣
∣
∣
∣
∣
∣
∣
4
1
2
3
5
9
6
7
1
∣
∣
∣
∣
∣
∣
∣
∣
=−199
D
x
=
∣
∣
∣
∣
∣
∣
∣
∣
3
5
9
6
7
1
25
13
1
∣
∣
∣
∣
∣
∣
∣
∣
=−796
D
y
=
∣
∣
∣
∣
∣
∣
∣
∣
4
25
6
1
13
7
2
1
1
∣
∣
∣
∣
∣
∣
∣
∣
=199 ,
D
z
=
∣
∣
∣
∣
∣
∣
∣
∣
4
3
25
1
5
13
2
9
1
∣
∣
∣
∣
∣
∣
∣
∣
=−398
Therefore x=
−199
−796
=4 , y=
−199
199
=−1 and z=
−199
−398
=2
Hence 4,−1,2 is the solution of the given simultaneous equations.
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