Question

Consider the system of equations
$x_1+2x_2+3x_3=1$
$x_2+2x_3+3x_1=2$
$x_3+2x_1+3x_2=3$
then

A. $x_1=\frac{2}{3}$

B. $x_1+x_2=\frac{4}{3}$

C.$x_3=-\frac{1}{3}$

D. $x_1-x_2=\frac{1}{3}$

Answer 1

For $x_1+2x_2+3x_3=1x_2+2x_3+3x_1=2x_3+2x_1+3x_2=3$
Let $M=\left[\begin{array}{cc}\begin{array}{ccc}1& 2& 3\\ 3& 1& 2\\ 2& 3& 1\end{array}& \begin{array}{c}1\\ 2\\ 3\end{array}\end{array}\right]$
Applying $R_2\to R_2-3R_1,R_3\to R_3-2R_1$
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& 2& 3\\ 0& -5& -7\\ 0& -1& -5\end{array}& \begin{array}{c}1\\ -1\\ 1\end{array}\end{array}\right]$
Dividing $R_2$ by $-5$, we get
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& 2& 3\\ 0& 1& \frac{7}{5}\\ 0& -1& -5\end{array}& \begin{array}{c}1\\ \frac{1}{5}\\ 1\end{array}\end{array}\right]$
Now applying $R_1\to R_1-2R_2,R_3\to R_3+R_2$
$M=\left[\begin{array}{cc}\begin{array}{ccc}1& 0& \frac{1}{5}\\ 0& 1& \frac{7}{5}\\ 0& 0& -\frac{18}{5}\end{array}& \begin{array}{c}\frac{2}{5}\\ \frac{7}{5}\\ \frac{6}{5}\end{array}\end{array}\right]$
Solving this we get $x_3=-\frac{1}{3},x_2=\frac{2}{3},x_1=\frac{2}{3}$
A) $x_1=\frac{2}{3}$
B) $x_1+x_2=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$
C) $x_3=-\frac{1}{3}$
D) $x_1-x_2=\frac{2}{3}-\frac{2}{3}=0$
Hence, options A, B and C are correct.