Question

Consider the following system of equation in three real variable $x_1,x_{2}and{x}_3$ :
$2x_1-x_2+3x_3=1$
$3x_1+2x_2+5x_3=2$
$-x_1+4x_2+x_3=3$
This system of equation has

• A. no solution
• B. a unique solution
• C. more than one but a finite number of solutions
• D. an infinite number of solutions

Answer 1

The correct option is B a unique solution

$[A:B]=\left[\begin{array}{ccccc}2& -1& 3& :& 1\\ 3& 2& 5& :& 2\\ -1& 4& 1& :& 3\end{array}\right]$

$\underset{\&R_1\leftrightarrow R_2}{\overset{R_3\to (-R_3)}{\to }}\left[\begin{array}{ccccc}3& 2& 5& :& 2\\ 2& -1& 3& :& 1\\ 1& -4& -1& :& -3\end{array}\right]$

$\stackrel{R_1\leftrightarrow R_3}{\to }\left[\begin{array}{ccccc}1& -4& -1& :& -3\\ 2& -1& 3& :& 1\\ 3& 2& 5& :& 2\end{array}\right]$
$R_2\to R_2-2R_1;R_3\to R_3-3R_1$

$\left[\begin{array}{c}A:B\end{array}\right]$ $\sim \left[\begin{array}{ccccc}1& -4& -1& :& -3\\ 0& 7& 5& :& 7\\ 0& 14& 8& :& 11\end{array}\right]$

$R_3\to R_3-2R_2$

$\left[\begin{array}{c}A:B\end{array}\right]$ $\sim \left[\begin{array}{ccccc}1& -4& -1& :& -3\\ 0& 7& 5& :& 7\\ 0& 0& -2& :& -3\end{array}\right]$

$\rho \left(A\right)=3=\rho \left(A|B\right)=3=\text{Number of unknowns.}$
$\therefore$ Unique soultion exists.