] Find the vector form of the general solution to the given linear system Ax = b. Then use that result to find the vector form of the general solution to Ax = 0.

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Find the vector form of the general solution to the given linear system Ax = b; Then use that result to find the vector form of the general solution to Ax = 0.

x1   + x2 + 2x3 = 5

x1            +  x3  = -2

2x1 + x2 +  3x3 = 3

the vector form of the general solution to the given linear system Ax = b

x1          -2                -1

x2   =     7         +s    -1

x3          0                 1

find the vector form of the general solution to Ax = 0.

x1                -1

x2   =    s    -1

x3                1

Step-by-step explanation:

Given that;

x1   + x2 + 2x3 = 5

x1            +  x3  = -2

2x1 + x2 +  3x3 = 3

Augmented matrix is expressed as;

1   1    2          5

1   0   3         -2

2  1    3          3

R2 ← R2 - R1        

1   1    2         5

0  -1   -1        -7

2  1    3         3

R3 ← R3 - 2R1

1   1    2         5

0  -1   -1        -7

0  -1   -1        -7

R3 ← R3 - R2

1   1    2         5

0  -1   -1        -7

0  0   0        0

R1 ← R1 + R2

1   0   1         -2

0  -1   -1        -7

0  0   0        0

R2 ← - R2

1   0   1         -2

0  1    1         7

0  0   0        0

SO from the above reduced matrix, we get

x1 = -2 - x3

x2 = 7 - x3

x3 = x3

now we introduce parameter "s" for free variable x3

x1 = -2 - s

x2 = 7 - s

x3 = s

x1          -2                -1

x2   =     7         +s    -1

x3          0                 1

General solution of Ax = 0 will be

x1                -1

x2   =    s    -1

x3                1

no he doesnt because "-16' is negatiive

step-by-step explanation: