Matrix 問題?

Question：
Using the Gaussian Elimination method, determine the values of a for which the linear system has no solutions, exactly one solution, or infinitely many solutions.
x + 2y + z = 2
2x - 2y + 3z = 1
x + 2y - az = a
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Solution：
Using the Gaussian Elimination method,
〔１　　２　　１｜２〕
〔２　—２　　３｜１〕
〔１　　２　—ａ｜ａ〕

　〔１　２　　　　１｜　　　　　　　　　　２〕
→〔０　１　—１／６｜　　　　　　　　１／２〕　(2R₁ - R₂)/6 → R₂
　〔０　０　　　　１｜（２—ａ）／（ａ＋１）〕　(R₁ - R₃)/(a + 1) → R₃

If the linear system has no solutions, a = -1.

If the linear system has exactly one solutions, a ≠ -1.

If the linear system has infinitely many solutions,
{ a = -1
{ a = 2
which is impossible.
This implies there is no value of a such that the linear system has infinitely many solutions.
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Remark：
If a ≠ -1,
x = (7a - 5)/[3(a + 1)]
y = (2a + 5)/[6(a + 1)]
z = -(a - 2)/(a + 1)