Question

Solve by Cramer's Rule: $x+y+z=2,$ $x+2y+z=1$ and $3x+y-5z=4$

• A. $x=\frac{5}{2},y=-1,z=\frac{1}{2}$
• B. $x=\frac{1}{2},y=-1,z=\frac{5}{2}$
• C. $x=\frac{5}{2},y=\frac{3}{2},z=\frac{1}{2}$
• D. $x=\frac{1}{2},y=\frac{3}{2},z=\frac{5}{2}$

Answer 1

The correct option is A $x=\frac{5}{2},y=-1,z=\frac{1}{2}$

Given equations are,

$x+y+z=2$

$x+2y+z=1$

$3x+y-5z=4$

Coefficient matrix $=\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\\ 3& 1& -5\end{array}\right]$

$△=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 1\\ 3& 1& -5\end{array}\right|=1(-10-1)-1(-5-3)+1(1-6)=-11+8-5=-8$

${△}_1=\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 4& 1& -5\end{array}\right|=2(-10-1)-1(-5-4)+1(1-8)=-22+9-7=-20$

${△}_2=\left|\begin{array}{ccc}1& 2& 1\\ 1& 1& 1\\ 3& 4& -5\end{array}\right|=1(-5-4)-2(-5-3)+1(4-3)=-9+16+1=8$

${△}_3=\left|\begin{array}{ccc}1& 1& 2\\ 1& 2& 1\\ 3& 1& 4\end{array}\right|=1(8-1)-1(4-3)+2(1-6)=7-1-10=-4$

$x=\frac{{△}_1}{△}=\frac{-20}{-8}=\frac{5}{2}$

$y=\frac{{△}_2}{△}=\frac{8}{-8}=-1$

$z=\frac{{△}_3}{△}=\frac{-4}{-8}=\frac{1}{2}$

$\therefore x=\frac{5}{2},y=-1$ and $z=\frac{1}{2}$

Hence solved.