solve the linear system by gauss-jordan elimination
− 2b + 3c = 1 ,
3a + 6b − 3c = −2,
6a + 6b + 3c = 5
The linear system can be written in matrix form as below
0 -2 3 a 1
3 6 -3 b = -2
6 6 3 c 5
we further reduce the above matrix as follows
The augmented matrix is
0 -2 3"\\mid"1
3 6 -3"\\mid" -2
6 6 3"\\mid" 5
applying R1"\\leftrightarrow" R3 we obtain
3 6 -3"\\mid" -2
0 -2 3"\\mid" 1
6 6 3"\\mid" 5
applying R3"\\to" R3-2R1 WE OBTAIN
3 6 -3"\\mid" -2
0 -2 3"\\mid" 1
0 -6 9"\\mid" 15
Applying R3"\\to" R3-3R2 we obtain
3 6 -3 "\\mid" -2
0 -2 3 "\\mid" 1
0 0 0 "\\mid" 12
applying R1"\\to" R1/3, R3"\\to" R3/12, R2"\\to" R2/-2, We obtain the below
1 2 -1 "\\mid" -2/3
0 1 -3/2 "\\mid" -1/2
0 0 0 "\\mid" 1
applying R1"\\to" R1-2R2 we obtain
1 0 2 "\\mid" 1/3
0 1 -3/2 "\\mid" -1/2
0 0 0 "\\mid" 1
using the above, we obtain the corresponding matrix as
a + 2c = 1/3
b - (3/2)c =-1/2
0a + 0b + 0c =1
The last equation gives 0=1 which is not possible.
Hence there is no solution.
Comments
Leave a comment