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.
Best Answer
$$ \begin{bmatrix} 1 & 1 & -1 & b_1 \\ 2 & -1 & 3 & b_2 \\ -1 & 3 & 1 & b_3 \\ 0 & 2 & -1 & b_4 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & -1 & b_1 \\ 0 & -3 & 5 & b_2 - 2 b_1 \\ 0 & 4 & 0 & b_1+b_3 \\ 0 & 2 & -1 & b_4 \end{bmatrix} $$ 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...