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Given the system $$ \begin{align*} -2x + ay - bz &= -4 \\ x + bz &= 2 \\ 2x + y + 3bz &= b \end{align*} $$ The question asks to find conditions on $a$ and $b$ that the system has no solutions, a unique solution or infinitely many solutions. If the system has many infinite solutions, what's solution in terms of $a$ and $b$.

I reduced the system to reduced echelon form. Now I'm stuck with it. Any help would be appreciated.

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    $\begingroup$ What's the reduced echelon form you found? $\endgroup$
    – egreg
    Jan 26 '15 at 8:38
  • $\begingroup$ i got x = b-2, ay + bz + 0, bz = b-4 $\endgroup$
    – jess_001
    Jan 26 '15 at 8:58
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Let $M=\begin{pmatrix} -2& a & -b\\ 1& 0 & b\\ 2&1&3b \end{pmatrix}$. Thus the system becomes: $$M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-4\\2\\b\end{pmatrix}.$$

In order for there to be a unique solution, $M$ must be invertible, i.e. $\det M\not=0$. Expanding the determinant along the middle row yields $\det M=-(3ab+b)-b(-2-2a)=b(1-a)$. Thus $b\not=0$ and $a\not=1$ guarantees a unique solution.

The remaining cases occur when $b=0$ or $a=1$. When $b=0$, we start by solving the bottom two rows:

$$\begin{pmatrix}1&0\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix}\implies\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2\\-4\end{pmatrix}.$$ Then to verify that the first row holds, we must have $(-2\ a)\cdot(2\ -4)=-4$, i.e. $a=2$. Since $z$ doesn't affect the solution, we have infinitely many solutions in this case: $$ a=2,\ b=0,\ t\in\mathbb{R}:\qquad \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\-4\\t\end{pmatrix} $$

Now suppose $a=1$. In this case, we see that $$0=(1\ 4\ -1)M\begin{pmatrix}x\\y\end{pmatrix}=(1\ 4\ -1)\begin{pmatrix}-4\\2\\b\end{pmatrix}=4-b$$ Thus when $b\not=4$, there are no solutions, and when $b=4$, there are infinitely many solutions. To see why, solve the bottom two rows: $$ \begin{pmatrix}1&0&4\\2&1&12\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\4\end{pmatrix}\implies \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-4t\\-4t\\t\end{pmatrix}. $$

Thus to summarize, when $a\not=1$ and $b\not=0$ there is a unique solution, and in the remaining cases here are the solutions: $$ (a,b)=\begin{cases} (2,0),& \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\-4\\t\end{pmatrix}\\ (1,4),& \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-4t\\-4t\\t\end{pmatrix}\\ \text{else},&\text{no solutions} \end{cases} $$

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Doing the elimination ($R_i$ is the $i$-th row)

  • exchange $R_2$ with $R_1$;
  • $R_2 \gets R_2+2R_1$;
  • $R_3 \gets R_3-2R_1$;
  • exchange $R_3$ with $R_2$;
  • $R_3 \gets R_3-aR_2$

we arrive at the form $$ \left[ \begin{array}{ccc|c} 1 & 0 & b & 2 \\ 0 & 1 & b & b-4 \\ 0 & 0 & b(1-a) & a(4-b) \end{array} \right] $$

  • If $b(1-a)\ne 0$, the system has a unique solution.

  • If $b=0$ and $a=0$, the system has infinitely many solutions

  • If $a=1$ and $b=4$, the system has infinitely many solutions

  • If $b=0$ and $a\ne0$ or $a=1$ and $b\ne4$, the system has no solution

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