Question

Solve the following sets of equations using Cramer's rule and remark about their consistency.

$x+2y+z=1$
$3x+y+z=6$
$x+2y=0$

• A. $x=-2,y=-1,z=-1$
• B. $x=2,y=-1,z=1$
• C. $x=-2,y=-1,z=1$
• D. $x=-2,y=1,z=1$

Answer 1

The correct option is B $x=2,y=-1,z=1$

Given,

$x+2y+z=1$
$3x+y+z=6$
$x+2y=0$

Using cramer's rule
$x=\frac{{\Delta }_1}{\Delta }=\frac{10}{5}=2,y=\frac{{\Delta }_2}{\Delta }=\frac{-5}{5}=-1,z=\frac{{\Delta }_3}{\Delta }=\frac{5}{5}=1$

$Explanation$

We know that,

$\Delta =\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ b_{11}& b_{12}& b_{13}\\ c_{11}& c_{12}& c_{13}\end{array}\right|$

$\Rightarrow$$\Delta =\left|\begin{array}{ccc}1& 2& 1\\ 3& 1& 1\\ 1& 2& 0\end{array}\right|=1(0-2)-2(0-1)+1(6-1)=-2+2+5=5$

${\Delta }_1=\left|\begin{array}{ccc}{d}_1& a_{12}& a_{13}\\ d_2& b_{12}& b_{13}\\ d_3& c_{12}& c_{13}\end{array}\right|$

$\Rightarrow$ ${\Delta }_1=\left|\begin{array}{ccc}1& 2& 1\\ 6& 1& 1\\ 0& 2& 0\end{array}\right|=1(0-2)-2(0-0)+1(12-0)=-2+12=10$

${\Delta }_2=\left|\begin{array}{ccc}{a}_{11}& d_1& a_{13}\\ b_{11}& d_2& b_{13}\\ c_{11}& d_3& c_{13}\end{array}\right|$

$\Rightarrow$ ${\Delta }_2=\left|\begin{array}{ccc}1& 1& 1\\ 3& 6& 1\\ 1& 0& 0\end{array}\right|=1(0-0)-1(0-1)+1(0-6)=0+1-6=-5$

${\Delta }_3=\left|\begin{array}{ccc}{a}_{11}& a_{12}& d_1\\ b_{11}& b_{12}& d_2\\ c_{11}& c_{12}& d_3\end{array}\right|$

$\Rightarrow {\Delta }_3=\left|\begin{array}{ccc}1& 2& 1\\ 3& 1& 6\\ 1& 2& 0\end{array}\right|=1(0-12)-2(0-6)+1(6-1)=-12+12+5=5$