By using properties of determinant prove that

 $\left|\begin{array}{ccc}\mathrm{x}+\mathrm{y}& \mathrm{y}+\mathrm{z}& \mathrm{z}+\mathrm{x}\\ \mathrm{z}& \mathrm{x}& \mathrm{y}\\ 1& 1& 1\end{array}\right|=0$

Without expanding determinant show that

$\left|\begin{array}{ccc}\mathrm{b}+\mathrm{c}& \mathrm{bc}& {\mathrm{b}}^2{\mathrm{c}}^2\\ \mathrm{c}+\mathrm{a}& \mathrm{ca}& {\mathrm{c}}^2{\mathrm{a}}^2\\ \mathrm{a}+\mathrm{b}& \mathrm{ab}& {\mathrm{a}}^2{\mathrm{b}}^2\end{array}\right|=0$

Without expanding determinant show that

$\left|\begin{array}{ccc}\mathrm{xa}& \mathrm{yb}& \mathrm{zc}\\ {\mathrm{a}}^2& {\mathrm{b}}^2& {\mathrm{c}}^2\\ 1& 1& 1\end{array}\right|=\left|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& \mathrm{z}\\ \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{bc}& \mathrm{ca}& \mathrm{ab}\end{array}\right|$

Without expanding determinant show that

$\left|\begin{array}{ccc}\mathrm{l}& \mathrm{m}& \mathrm{n}\\ \mathrm{e}& \mathrm{d}& \mathrm{f}\\ \mathrm{u}& \mathrm{v}& \mathrm{w}\end{array}\right|=\left|\begin{array}{ccc}\mathrm{n}& \mathrm{f}& \mathrm{w}\\ \mathrm{l}& \mathrm{e}& \mathrm{u}\\ \mathrm{m}& \mathrm{d}& \mathrm{v}\end{array}\right|$

Without expanding determinant show that

$\left|\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}\right|=0$

If$\left|\begin{array}{ccc}\mathrm{a}& 1& 1\\ 1& \mathrm{b}& 1\\ 1& 1& c\end{array}\right|=0$ then show that $\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=1$

Solve the following linear equations by Cramer’s Rule.

$2\mathrm{x}-\mathrm{y}+\mathrm{z}=1,\mathrm{x}+2\mathrm{y}+3\mathrm{z}=8,3\mathrm{x}+\mathrm{y}-4\mathrm{z}=1$

Solve the following linear equations by Cramer’s Rule.

$\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{3}{2}, \frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=\frac{5}{6},\frac{1}{\mathrm{z}}+\frac{1}{\mathrm{x}}=\frac{4}{3}$

Solve the following linear equations by Cramer’s Rule.

$2\mathrm{x}+3\mathrm{y}+3\mathrm{z}=5,\mathrm{x}-2\mathrm{y}+\mathrm{z} =–4,$$3 \mathrm{x}–\mathrm{y}–2\mathrm{z}=3$