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Expert-verified Found in: Page 39 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Question: Solve the linear system $|\begin{array}{cc}& \text{}y+z=a\\ X& +z=b\\ x+y& =C\end{array}|$, where a,b and c are arbitrary constants.

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

See the step by step solution

## Step 1: 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.

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

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, . ### Want to see more solutions like these? 