We can set up a matrix and use Gaussian elimination .
( 1 2 − 1 0 3 1 1 − 5 3 ) \begin{pmatrix}
1 & 2 & -1 \\
0 & 3 & 1 \\
1 & -5 & 3 \\
\end{pmatrix} ⎝ ⎛ 1 0 1 2 3 − 5 − 1 1 3 ⎠ ⎞ R 3 = R 3 − R 1 R_3=R_3-R_1 R 3 = R 3 − R 1
( 1 2 − 1 0 3 1 0 − 7 4 ) \begin{pmatrix}
1 & 2 & -1 \\
0 & 3 & 1 \\
0 & -7 & 4 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 2 3 − 7 − 1 1 4 ⎠ ⎞ R 2 = R 2 / 3 R_2=R_2/3 R 2 = R 2 /3
( 1 2 − 1 0 1 1 / 3 0 − 7 4 ) \begin{pmatrix}
1 & 2 & -1 \\
0 & 1 & 1/3 \\
0 & -7 & 4 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 2 1 − 7 − 1 1/3 4 ⎠ ⎞ R 1 = R 1 − 2 R 2 R_1=R_1-2R_2 R 1 = R 1 − 2 R 2
( 1 0 − 5 / 3 0 1 1 / 3 0 − 7 4 ) \begin{pmatrix}
1 & 0 & -5/3 \\
0 & 1 & 1/3 \\
0 & -7 & 4 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 − 7 − 5/3 1/3 4 ⎠ ⎞ R 3 = R 3 + 7 R 2 R_3=R_3+7R_2 R 3 = R 3 + 7 R 2
( 1 0 − 5 / 3 0 1 1 / 3 0 0 19 / 3 ) \begin{pmatrix}
1 & 0 & -5/3 \\
0 & 1 & 1/3 \\
0 & 0 & 19/3\\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 − 5/3 1/3 19/3 ⎠ ⎞ R 3 = ( 3 / 19 ) R 3 R_3=(3/19)R_3 R 3 = ( 3/19 ) R 3
( 1 0 − 5 / 3 0 1 1 / 3 0 0 1 ) \begin{pmatrix}
1 & 0 & -5/3 \\
0 & 1 & 1/3 \\
0 & 0 & 1\\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 − 5/3 1/3 1 ⎠ ⎞ R 1 = R 1 + ( 5 / 3 ) R 3 R_1=R_1+(5/3)R_3 R 1 = R 1 + ( 5/3 ) R 3
( 1 0 0 0 1 1 / 3 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 1/3 \\
0 & 0 & 1\\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 1/3 1 ⎠ ⎞ R 2 = R 2 − R 3 / 3 R_2=R_2-R_3/3 R 2 = R 2 − R 3 /3
( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1\\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ Since the rank of the matrix is 3, then the set {(1,2,-1),(0,3,1),(1,-5,3)} is a basis of R 3 . R^3. R 3 .
The dimension is 3.
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