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What conditions must be satisfied for the overdetermined system below to be consistent? Do not try to solve for the solution naively. Apply row reduction. If the system is consistent does a unique solution exist?

$x_1 + x_2 - x_3 = b_1$
$2x_1 - x_2 + 3x_3 = b_2$
$-x_1 + 3x_2 + x_3 = b_3$
$2x_2 - x_3 = b_4$

I did apply row reduction and found 4th row all zeros (of course its because overdetermined and equations>variables) and still don't know whether it has unique solution or not.

Please someone explain it with simple(as much as possible) English words because my English is limited.

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  • $\begingroup$ Could you include what you obtained from the row reduction? $\endgroup$ Commented Dec 26, 2013 at 17:05
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    $\begingroup$ You didn't find the fourth row completely zero. The right hand side wasn't. $\endgroup$
    – OR.
    Commented Dec 26, 2013 at 17:05
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    $\begingroup$ Here the calculation is lengthy. It is not a problem. Find out the row reduced augmented matrix for $AX=B$ i.e. $[A,B]$ and see the element of the last row and last column. $\endgroup$
    – Supriyo
    Commented Dec 26, 2013 at 17:17

1 Answer 1

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$$ \begin{bmatrix} 1 & 1 & -1 & b_1 \\ 2 & -1 & 3 & b_2 \\ -1 & 3 & 1 & b_3 \\ 0 & 2 & -1 & b_4 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & -1 & b_1 \\ 0 & -3 & 5 & b_2 - 2 b_1 \\ 0 & 4 & 0 & b_1+b_3 \\ 0 & 2 & -1 & b_4 \end{bmatrix} $$ Now continue row-reduction until you end up with the last row of 3 zeroes and some expression with $b_1, \ldots, b_4$ on the right-hand side. Call this expression $f$.

Note that if $f \neq 0$, the system has no solutions. However, if indeed $x=0$, then you can read off the correct values for $x_1,x_2,x_3$ from the reduced form of the matrix that gives your unique solution, provided also you don't divide by zero...

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  • $\begingroup$ you mean $f \neq 0$ is all element in 4th column are not zero ? $\endgroup$
    – Andy
    Commented Dec 26, 2013 at 17:43
  • $\begingroup$ @Andy No. The 4th row has the first 3 entries of 0 and the last one is $f$. The last equation reads $0 x_1 + 0 x_2 + 0 x_3 = f$, so if $f \neq 0$, there are no solutions. $\endgroup$
    – gt6989b
    Commented Dec 26, 2013 at 18:41

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