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As the title says. I have these linear systems:

$x_1 + 3x_2 - x_3 + 2x_4 = b_1$

$-2x_1 + x_2 +5x_3 + x_4 = b_2$

$3x_1 - 2x_2 - 2x_3 + x_4 = b_3$

$5x_1 - 7x_2 - 3x_3 = b_4$

I need to determine the conditions of the b's. I tried to solve this by using gauss elimination to convert it into echelon form, but that seems unnecessarily cumbersome to solve this, thanks to be b's. I was hoping there would be an easier way to do this.

The correct answers are: $b_1 -2b_3 + b_4 = 0$ or $b_1 = 2b_3 - b_4$

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  • $\begingroup$ Could I perhaps see your calculation steps? How do you end up with only zeroes at the bottom? I end up with with something that looks really wrong: $x3 = -(176b3/(-155)) + (208b1/(-155))+(64b4/(-155))$ $\endgroup$
    – Elias
    Commented Nov 28, 2017 at 23:04

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For the system to be consistent, $(b_1,b_2,b_3,b_4)^T$ must be an element of the column space of the coefficient matrix. Gaussian elimination is the way to go, but you don’t need to carry the $b$’s along in an augmented matrix. Reducing the transpose of the coefficient matrix (i.e., column-reducing it) will give you a “nice” basis for the column space from which you can read off constraints on the $b$’s.

Specifically, perform the reduction $$\begin{bmatrix}1&-2&3&5\\3&1&-2&-7\\-1&5&-2&-3\\2&1&1&0\end{bmatrix} \to \begin{bmatrix}1&0&0&-1\\0&1&0&0\\0&0&1&2\\0&0&0&0\end{bmatrix}.$$ From this you can see that the first three $b$’s are free, while the last column gives the constraint $b_4=-b_1+2b_3$.


Update: Per request, I’m adding some of the intermediate stages of the row-reduction. The upper-left entry is already $1$, so just clear the first column: $$\begin{bmatrix}1&-2&3&5\\0&7&-11&-22\\0&3&1&2\\0&5&-5&-10\end{bmatrix}$$ Looking now at the second column, dividing by $7$ or $3$ looks slightly messy, but diving the last row by $5$ looks like it’ll keep things tidier, so do that, swap it with the second row, and clear the rest of the column: $$\begin{bmatrix}1&0&1&1\\0&1&-1&-2\\0&0&4&8\\0&0&-4&-8\end{bmatrix}$$ At this point, it’s obvious that the matrix has rank 3 and we have a basis for the row space (remember, that’s the column space of the original coefficient matrix), but we can simplify further. Divide the third row by $4$ and clear the third column: $$\begin{bmatrix}1&0&0&-1\\0&1&0&0\\0&0&1&2\\0&0&0&0\end{bmatrix}$$ Vectors in the column space of the original coefficient matrix are therefore of the form $(x_1,x_2,x_3,2x_3-x_1)^T$, whence the constraint $b_4=2b^3-b_1$.

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