Assume the general solution of the augmented matrix [1 2 3 -1] [4 5 6 5] [7 8 9

Question

Assume the general solution of the augmented matrix

[1 2 3 -1]
[4 5 6 5]
[7 8 9 11]

is as follows:

x1=5+x3
x2=-3-2x3
x3 is free


Using the three properties of a subspace, explain why or whynot this solution forms a subspace of R3.


The three properties are as follows:

1) the zero vector is in the set
2) for each u and v in the set, the sum u+v is in the set (i.e.,the set is closed under addition)
3) for each u in the set and each scalar c, the vector cu is in theset (i.e., the set is closed under scalar multiplication).

Thanks! =) Assume the general solution of the augmented matrix

[1 2 3 -1]
[4 5 6 5]
[7 8 9 11]

is as follows:

x1=5+x3
x2=-3-2x3
x3 is free


Using the three properties of a subspace, explain why or whynot this solution forms a subspace of R3.


The three properties are as follows:

1) the zero vector is in the set
2) for each u and v in the set, the sum u+v is in the set (i.e.,the set is closed under addition)
3) for each u in the set and each scalar c, the vector cu is in theset (i.e., the set is closed under scalar multiplication).

Thanks! =)

Explanation / Answer

The zero vector is not in the solution space. If you set x3 =0, then your equation say x2 must be -3 and x1 must be 5. So(0, 0, 0) is not in your solution space, so it fails condition 1)to be a subspace.

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