Question

Question:Solve the linear system

$\left|\begin{array}{cc}& y+z=a\\ X& +z=b\\ x+y& =C\end{array}\right|$, where a,b andcare arbitrary constants.

Short answer

Answer:

The solution of the linear system is $\left|\begin{array}{c}y+z=a\\ y+z=b\\ y+y=c\end{array}\right|,x=\frac{b+c-a}{2},y=\frac{a+c-b}{2},z=\frac{a+b-c}{2}$,

Step-by-step solution

Consider the system

Consider the linear system.

$\begin{array}{c}y+z=a\\ y+z=b\\ y+y=c\end{array}$

The matrix form of the system is,

$\left[\begin{array}{ccccc}0& 1& 1& |& 1\\ 1& 0& 1& |& b\\ 1& 1& 0& |& c\end{array}\right]$

 Step 2: Compute the system.

Consider the matrix,

$\left[\begin{array}{ccccc}0& 1& 1& |& 1\\ 1& 0& 1& |& b\\ 1& 1& 0& |& c\end{array}\right]$

Perform the row reduction operation.

$$\begin{gathered} \left[\begin{array}{ccccc}0& 1& 1& |& 1\\ 1& 0& 1& |& b\\ 1& 1& 0& |& c\end{array}\right]~\left[\begin{array}{ccccc}0& 1& 1& & 1\\ 1& 0& 1& & b\\ 1& 1& -2& & c-b-a\end{array}\right] \\ \left[\begin{array}{cccc}1& 0& 0& \frac{b-a+c}{2}\\ 0& 1& 0& \frac{a-b+c}{2}\\ 0& 0& 1& \frac{a+b-c}{2}\end{array}\right] \end{gathered}$$

The values are, $x=\frac{b+c-a}{2},y=\frac{a+c-b}{2},z=\frac{a+b-c}{2}$

Hence,$\left|\begin{array}{c}y+z=a\\ y+z=b\\ y+y=c\end{array}\right|,$ $,x=\frac{b+c-a}{2},y=\frac{a+c-b}{2},z=\frac{a+b-c}{2}$ is the solution of the linear system, .