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Consider the following system of linear equations: \begin{eqnarray} { {x_1 + 2x_2 - x_3 = 1 }\\ {x_2 + \alpha x_3 = 0} \\ {2x_1 + 2x_2 + x_3 = \beta}} \end{eqnarray}

I have the row reduced matrix below but I end up with a pivot position in every row so I'm not sure how to fine values of $\alpha$ and $\beta$ that give infinite solutions.

\begin{bmatrix} 1 & 0 & -1-2 \alpha & 1 \\ 0 & 1 & \alpha & 0 \\ 0 & 0 & 1 & \frac{\beta - 1}{2} \end{bmatrix}

Please let me know what I'm doing/I've done wrong.

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2 Answers 2

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Third equation $-$ first gives $$x_1=\beta-1$$

first equation $-2$ second gives $$(1-2\alpha)x_3=2-\beta $$

thus we need $$\alpha=\frac {1}{2}\;,\;\beta=2$$

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  • $\begingroup$ Sorry I'd written the first equation wrong, edited now. $\endgroup$
    – COvert96
    Jun 27, 2017 at 10:44
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By elimination, the $2^{nd}$ equation gives $x_2=-\alpha x_3\,$, and substituting back into the $1^{st}$ equation gives $x_1=1-2(-\alpha x_3)+x_3=1+(2 \alpha+1)x_3\,$. Then the $3^{rd}$ equation reduces to:

$$ 2\big(1+(2 \alpha+1)x_3\big) + 2(-\alpha x_3) + x_3 = \beta \quad\iff\quad (2 \alpha +3)x_3 = \beta-2 $$

The latter will have a unique solution in $x_3$ unless $2 \alpha+3=0\,$, and will have infinitely many solutions iff $2 \alpha+3=\beta-2=0\,$.

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