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In order to find values of b and c such that the system $x_1 + x_2 + bx_3 =1$, $bx_1 + 3x_2 - x_3 = -2$, $3x_1 + 4x_2 + x_3 = c$ contains no solution, we have to find values for b and c such that the augmented matrix form of this system of equations contains a row that has a 0 on the left hand side, and some value $\neq 0$ on the right hand side.

$ \left( \begin{array}{ccc|c} 1&1&b&1\\b&3&-1&-2\\3&4&1&c \end{array} \right) $ $\to$ $ \left( \begin{array}{ccc|c} 1&1&b&1\\0&3-b&-1-b^2&-2-b\\0&1&1-3b&c-3 \end{array} \right) $

What should I do next? If I try to set $3-b = -1$, $-1 - b^2 = 1-3b$ I get b = 4 but that doesn't set any of the left hand rows to 0.

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    $\begingroup$ determinant=0 at first. $\endgroup$ – hamam_Abdallah Oct 15 '16 at 23:04
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if we compute the determinant, we will find

$D=4b^2-10b+4$

$D=0$ if $b=2$ or $b=\frac{1}{2}$

now if $b=2$

then the system becomes

$x+y+2z=1$

$2x+3y-z=-2$

$3x+4y+z=c$

the sum of the two first equations give

$3x+4y+z=-1$

so if $c\neq -1$ there will be no solution.

now if $b=\frac{1}{2}$

the system becomes

$2x+2y+z=2$

$x+6y-2z=-4$

$3x+4y+z=c$

$5\times $third$-7\times$ first gives

$x+6y-2z=5c-14$

so if $5c-14\neq -4$ , or in other terms

$c\neq 2$, there is no solution.

finally, the system has no solution if

$b=2$ and $c\neq -1$ or

$b=0.5$ and $c\neq 2$.

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  • $\begingroup$ Thanks, one more question. What do I do if the problem asks for exactly one solution? That means that I have to find b and c such that there are an equal number of pivots as there are rows, right? $\endgroup$ – okmanl Oct 16 '16 at 0:00
  • $\begingroup$ there is one solution if determinant $\neq 0$ $\endgroup$ – hamam_Abdallah Oct 16 '16 at 0:07

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