# Solve the following sets of equations using Cramer's rule and remark about theit consistency. (a) x + y + z - 6 = 0     2x + y - z - 1 = 0     x + y - 2z + 3 = 0 (b) x + 2y + z = 1     3x + y + z = 6     x + 2y = 0 (c) x - 3y + z = 2     3x + y + z = 6     5x + y + 3z = 3 (d) 7x - 7y + 5z = 3     3x + y + 5z = 7    2x + 3y + 5z = 5

Dear Student,

$\mathrm{x}+\mathrm{y}+\mathrm{z}=6\phantom{\rule{0ex}{0ex}}2\mathrm{x}+\mathrm{y}-\mathrm{z}=1\phantom{\rule{0ex}{0ex}}\mathrm{x}+\mathrm{y}-2\mathrm{z}=-3\phantom{\rule{0ex}{0ex}}\mathrm{D}=\left|\begin{array}{ccc}1& 1& 1\\ 2& 1& -1\\ 1& 1& -2\end{array}\right|=1\left(-2+1\right)-1\left(-4+1\right)+1\left(2-1\right)=-1+3+1=3\phantom{\rule{0ex}{0ex}}{\mathrm{D}}_{1}=\left|\begin{array}{ccc}6& 1& 1\\ 1& 1& -1\\ -3& 1& -2\end{array}\right|=6\left(-2+1\right)-1\left(-2-3\right)+1\left(1+3\right)=-6+5+4=3\phantom{\rule{0ex}{0ex}}{\mathrm{D}}_{2}=\left|\begin{array}{ccc}1& 6& 1\\ 2& 1& -1\\ 1& -3& -2\end{array}\right|=1\left(-2-3\right)-6\left(-4+1\right)+1\left(-6-1\right)=-5+18-7=6\phantom{\rule{0ex}{0ex}}{\mathrm{D}}_{3}=\left|\begin{array}{ccc}1& 1& 6\\ 2& 1& 1\\ 1& 1& -3\end{array}\right|=1\left(-3-1\right)-1\left(-6-1\right)+6\left(2-1\right)=-4+7+6=9\phantom{\rule{0ex}{0ex}}\mathrm{x}=\frac{{\mathrm{D}}_{1}}{\mathrm{D}}=\frac{3}{3}=1\phantom{\rule{0ex}{0ex}}\mathrm{y}=\frac{{\mathrm{D}}_{2}}{\mathrm{D}}=\frac{6}{3}=2\phantom{\rule{0ex}{0ex}}\mathrm{z}=\frac{{\mathrm{D}}_{3}}{\mathrm{D}}=\frac{9}{3}=3\phantom{\rule{0ex}{0ex}}$

Regards

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