Question

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

$x-3y+z=2$
$3x+y+z=6$
$5x+y+3z=3$

• A. $x=\frac{13}{3},y=-\frac{7}{6},z=-\frac{35}{6}$;
• B. $x=\frac{-13}{3},y=\frac{7}{6},z=-\frac{35}{6}$;
• C. $x=\frac{-13}{3},y=\frac{7}{6},z=\frac{35}{6}$;
• D. $x=\frac{13}{3},y=\frac{7}{6},z=\frac{35}{6}$;

Answer 1

The correct option is A $x=\frac{13}{3},y=-\frac{7}{6},z=-\frac{35}{6}$;

Given,

$x-3y+z=2$
$3x+y+z=6$
$5x+y+3z=3$

Using Cramer's rule
$x=\frac{{\Delta }_1}{\Delta }=\frac{52}{12}=\frac{13}{3},y=\frac{{\Delta }_2}{\Delta }=\frac{-14}{12}=-\frac{7}{6},z=\frac{{\Delta }_3}{\Delta }=\frac{-70}{12}=-\frac{35}{6}$

We have,

$a_{11}=1,a_{12}=-3,a_{13}=1,b_{11}=3,b_{12}=1,b_{13}=1,c_{11}=5,c_{12}=1,c_{13}=3d_{11}=2,d_{12}=6,d_{13}=3$

By cramer's rule we have,

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

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

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

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

$\Delta =\left|\begin{array}{ccc}1& -3& 1\\ 3& 1& 1\\ 5& 1& 3\end{array}\right|=1(3-1)+3(9-5)+1(3-5)=2+12-2=12{\Delta }_1=\left|\begin{array}{ccc}2& -3& 1\\ 6& 1& 1\\ 3& 1& 3\end{array}\right|=2(3-1)+3(18-3)+1(6-3)=4+45+3=52{\Delta }_2=\left|\begin{array}{ccc}1& 2& 1\\ 3& 6& 1\\ 5& 3& 3\end{array}\right|=1(18-3)-2(9-5)+1(9-30)=15-8-21=-14{\Delta }_3=\left|\begin{array}{ccc}1& -3& 2\\ 3& 1& 6\\ 5& 1& 3\end{array}\right|=1(3-6)+3(9-30)+2(3-5)=-3-63-4=-70$