Question

The number of solutions of the system of equations: $2x+y-z=7$, $x-3y+2z=1$, $x+4y-3z=5$, is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Given,

$2x+y-z=7$

$x-3y+2z=1$

$x+4y-3z=5$

Arranging the above equations in form of matrix and finding the determinant, we get,

$\Delta =\left|\begin{array}{ccc}2& 1& -1\\ 1& -3& 2\\ 1& 4& -3\end{array}\right|$

$=2(9-8)-1(-3-2)-1(4+3)$

$\therefore \Delta =2+5-7=0$

Now,

${\Delta }_1=\left|\begin{array}{ccc}7& 1& -1\\ 1& -3& 2\\ 5& 4& -3\end{array}\right|$

$=7(9-8)-1(-3-10)-1(4+15)$

$\therefore {\Delta }_1=7+13-19=1\ne 0$

Hence the given system of equations does not have any solution.