We can set up a matrix and use Gaussian elimination .
⎝⎛10123−5−113⎠⎞ R3=R3−R1
⎝⎛10023−7−114⎠⎞R2=R2/3
⎝⎛10021−7−11/34⎠⎞R1=R1−2R2
⎝⎛10001−7−5/31/34⎠⎞R3=R3+7R2
⎝⎛100010−5/31/319/3⎠⎞ R3=(3/19)R3
⎝⎛100010−5/31/31⎠⎞ R1=R1+(5/3)R3
⎝⎛10001001/31⎠⎞ R2=R2−R3/3
⎝⎛100010001⎠⎞ Since the rank of the matrix is 3, then the set {(1,2,-1),(0,3,1),(1,-5,3)} is a basis of R3.
The dimension is 3.
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