Question

solve the linear system by gauss-jordan elimination

 − 2b + 3c = 1 ,

3a + 6b − 3c = −2,

6a + 6b + 3c = 5

Answer 1

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.