Answer to Question #298577 in Linear Algebra for shahana

Question #298577

solve the linear system by gauss-jordan elimination

 − 2b + 3c = 1 ,

3a + 6b − 3c = −2,

6a + 6b + 3c = 5


1
Expert's answer
2022-02-17T10:53:04-0500

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.



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