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For which values of $a$ does the system $$x_1 + x_2 + x_3 = 1$$ $$x_1 + 2x_2 + ax_3 = 2$$ $$2x_1 + ax_2 + 4x_3 = a^2$$ have (i) a unique solution, (ii) no solution, (iii) infinitely many solutions? Where the system has infinitely many solutions, write the solutions in parametric form.

So I tried to row reduce the matrix and got up till this point:

$$\left[\begin{array}{ccc|c}1&1&1&1\\ 0&1&a-1&1\\ 0&a-2&2&a^2-2\end{array}\right]$$

But I'm a little confused on how to continue further. How exactly would I change the $a-2$ to a $0$ and $2$ to a $1$?

Any help?

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    $\begingroup$ Subtract the second row times $a-2$ from the third. Then divide the third row by the element $3,3$. $\endgroup$ – Yves Daoust Dec 12 '17 at 15:57
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$\det \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & a \\ 2 & a & 4 \\ \end{array} \right)=3 a-a^2$

$3a-a^2\ne 0\to a\ne 0\lor a\ne 3$ the sistem has one and only one solution

If $a=0$ the system becomes

$\begin{cases} x_1 + x_2 + x_3=1\\ x_1 + 2 x_2=2\\ 2 x_1 + 4 x_3=0\\ \end{cases} $

Infinite solutions $(t,1-t/2,-t/2)$

if $a=3$ then $\text{rank}(A)=2$

$ A|B=\left( \begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 2 \\ 2 & 3 & 4 & 9 \\ \end{array} \right)$

$\text{rank}(A|B)=3\ne \text{rank}(A)$

so the system is impossible

Hope this helps

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Hint:

The discussion depends on the zeroes of the main determinant. Developing with the first column,

$$\Delta=2-(a-2)(a-1).$$

The rest is yours.

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You are on the right track :)

So you have \begin{align} &\left[\begin{array}{ccc|c} 1&1&1&1\\ 0&1&a-1&1\\ 0&a-2&2&a^2-2 \end{array}\right]\\ \xrightarrow{R_3-(a-2)R_2\to R_3}\ &\left[\begin{array}{ccc|c} 1&1&1&1\\ 0&1&a-1&1\\ 0&0&2-(a-2)(a-1)&(a^2-2)-(a-2)\\ \end{array}\right]\\ =&\left[\begin{array}{ccc|c} 1&1&1&1\\ 0&1&a-1&1\\ 0&0&a(3-a)&a(a-1)\\ \end{array}\right]\\ \end{align}

Can you proceed from here?

Consider the cases where $a=0,3$.

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  • $\begingroup$ Why should you consider $a=1$ ? $\endgroup$ – Yves Daoust Dec 12 '17 at 16:17
  • $\begingroup$ @YvesDaoust don't we have inconsistent system with $a=1$ or did I make some careless mistake? $\endgroup$ – Karn Watcharasupat Dec 12 '17 at 16:43
  • $\begingroup$ When $a=1$, the unique solution is $x_1=0,x_2=1,x_3=0$. $\endgroup$ – Yves Daoust Dec 12 '17 at 16:55

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