System of Equation

Question

Solve the system of equations

$$-x_1+x_2+x_3=a$$ $$x_1-x_2+x_3=b$$ $$x_1+x_2-x_3=c$$

I have tried writing the augmented matrix of the system of equations above and reducing it into echelon form but that didn't work out. Please help.

Answer 1

Adding the first two equations, $$2x_{3}=a+b$$

Can you proceed in the similar manner?

Answer 2

For a $3\times 3$ system with only $1$ and $-1$ coefficients, it's easier to isolate one unknown from one equation and to replace it in an other equation. Here for example equation $3$ gives you : $x_3 = x_1 + x_2 - c$. Then you can replace $x_3$ in equation $2$, isolate $x_2$ and replace in eq. $1$.

Answer 3

Let's do the Gaussian elimination: \begin{align} \left[\begin{array}{ccc|c} -1 & 1 & 1 & a \\ 1 & -1 & 1 & b \\ 1 & 1 & -1 & c \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & -1 & -1 & -a \\ 1 & -1 & 1 & b \\ 1 & 1 & -1 & c \end{array}\right] && R_1\gets -R_1 \\ &\to \left[\begin{array}{ccc|c} 1 & -1 & -1 & -a \\ 0 & 0 & 2 & b+a \\ 0 & 2 & 0 & c+a \end{array}\right] && \begin{aligned}R_2&\gets R_2-R_1\\R_3&\gets R_3-R_1\end{aligned} \\ &\to \left[\begin{array}{ccc|c} 1 & -1 & -1 & -a \\ 0 & 2 & 0 & c+a \\ 0 & 0 & 2 & b+a \end{array}\right] && R_2\leftrightarrow R_3 \\ &\to \left[\begin{array}{ccc|c} 1 & -1 & -1 & -a \\ 0 & 1 & 0 & (c+a)/2 \\ 0 & 0 & 1 & (b+a)/2 \end{array}\right] && \begin{aligned}R_2&\gets \tfrac{1}{2}R_2\\R_3&\gets \tfrac{1}{2}R_3\end{aligned} \\ &\to \left[\begin{array}{ccc|c} 1 & -1 & 0 & (b-a)/2 \\ 0 & 1 & 0 & (c+a)/2 \\ 0 & 0 & 1 & (b+a)/2 \end{array}\right] &&R_1\gets R_1+R_3 \\ &\to \left[\begin{array}{ccc|c} 1 & 0 & 0 & (b+c)/2 \\ 0 & 1 & 0 & (c+a)/2 \\ 0 & 0 & 1 & (b+a)/2 \end{array}\right] &&R_1\gets R_1+R_2 \end{align}