Let us find the rank of the extended matrix of the system of equations:
( 1 − 1 1 1 1 − 2 2 0 − 1 ∣ 1 2 3 ) − I − 2 I ∼ ( 1 − 1 1 0 2 − 3 0 2 − 3 ∣ 1 1 1 ) − I I ∼ ( 1 − 1 1 0 2 − 3 0 0 0 ∣ 1 1 0 ) \left( {\left. {\begin{matrix}
1&{ - 1}&1\\
1&1&{ - 2}\\
2&0&{ - 1}
\end{matrix}} \right|\begin{matrix}
1\\
2\\
3
\end{matrix}} \right)\begin{matrix}
{}\\
{ - I}\\
{ - 2I}
\end{matrix} \sim \left( {\left. {\begin{matrix}
1&{ - 1}&1\\
0&2&{ - 3}\\
0&2&{ - 3}
\end{matrix}} \right|\begin{matrix}
1\\
1\\
1
\end{matrix}} \right)\begin{matrix}
{}\\
{ - II}
\end{matrix} \sim \left( {\left. {\begin{matrix}
1&{ - 1}&1\\
0&2&{ - 3}\\
0&0&0
\end{matrix}} \right|\begin{matrix}
1\\
1\\
0
\end{matrix}} \right) ⎝ ⎛ 1 1 2 − 1 1 0 1 − 2 − 1 ∣ ∣ 1 2 3 ⎠ ⎞ − I − 2 I ∼ ⎝ ⎛ 1 0 0 − 1 2 2 1 − 3 − 3 ∣ ∣ 1 1 1 ⎠ ⎞ − II ∼ ⎝ ⎛ 1 0 0 − 1 2 0 1 − 3 0 ∣ ∣ 1 1 0 ⎠ ⎞
Since the ranks of the main and extended matrix are equal to each other, but not equal to the number of equations:
r a n k ( A ) = r a n k ( A ∣ B ) = 2 ≠ 3 rank(A) = rank(A|B) = 2 \ne 3 r ank ( A ) = r ank ( A ∣ B ) = 2 = 3 , the system of equations is consistent, but not defined, and therefore has an infinite number of solutions
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