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\begin{align} x_1 + x_2 + cx_3 &= 0 \\ x_2 + x_3 &= 0 \\ cx_1 + x_3 &= 1 \end{align}

I just started learning about linear systems and I'm really confused on how to start on this problem. I'm trying to make a row to be $\begin{pmatrix} 0 & 0 & 0 &=& 1 \end{pmatrix}$ so that its inconsistent but I'm not sure where to start.

Thanks a lot!!

Anon
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geeeeeek
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  • Maybe try elimination. Or substitute $x_2=-x_3$ into the first so that if you look at the first and third equation, you now see a system of $2$ equations in two unknowns. – Ahmed S. Attaalla Oct 04 '18 at 20:17

3 Answers3

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For a inconsistent system, $$ \begin{vmatrix} 1 & 1 & c \\ 0 & 1 & 1 \\ c & 0 & 1 \end{vmatrix} = 0 $$ and atleast one of $ \begin{vmatrix} 0 & 1 & c \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix}$, $ \begin{vmatrix} 1 & 0 & c \\ 0 & 0 & 1 \\ c & 1 & 1 \end{vmatrix}$ or $ \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ c & 0 & 1 \end{vmatrix}$ must be non-zero.

Anon
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Let's compute a row echelon form of the system’s matrix \begin{align} \begin{bmatrix} 1 & 1 & c & 0 \\ 0 & 1 & 1 & 0 \\ c & 0 & 1 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & c & 0 \\ 0 & 1 & 1 & 0 \\ 0 & -c & 1-c^2 & 1 \end{bmatrix} && R_3\gets R_3-cR_1 \\[6px]&\to \begin{bmatrix} 1 & 1 & c & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1+c-c^2 & 1 \end{bmatrix} && R_3\gets R_3+cR_2 \end{align} The system is inconsistent if and only if $1+c-c^2=0$.

You might also try and solve the system by noticing that $x_2=-x_3$ and $x_3=1-cx_1$, so the first equation becomes $$ x_1-(1-cx_1)+c(1-cx_1)=0 $$ or $$ (1+c-c^2)x_1=1-c $$

egreg
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Let's row reduce the augmented matrix associated with the system: $\left(\begin{array}{rrr|r}1&1&c&0\\0&1&1&0\\c&0&1&1\end{array}\right)\to\left(\begin{array}{rrr|r}1&1&c&0\\0&1&1&0\\0&-c&1-c^2&1\end{array}\right)\to\left(\begin{array}{rrr|r}1&1&c&0\\0&1&1&0\\0&0&1+c-c^2&1\end{array}\right)$.

So to answer the question, $1+c-c^2=0$ gives an inconsistent system. So $c=\frac{1\pm\sqrt5}2$.