Find the product of the matrix $\mathrm{A}=\left[\begin{array}{ccc}2& 3& 4\\ 5& 4& -6\\ 3& -2& -2\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}20& 2& 34\\ 8& 16& -32\\ 22& -13& 7\end{array}\right]$, Hence, solve the system of equations:

$\frac{2}{x}+\frac{3}{y}+\frac{{\displaystyle 4}}{z}=-3, \frac{5}{x}+\frac{4}{y}-\frac{{\displaystyle 6}}{z}=4, \frac{3}{x}-\frac{2}{y}-\frac{{\displaystyle 2}}{z}=6$.

If $\mathrm{A}$ and $\mathrm{B}$ are square matrix of order $3$ such that $\left|\mathrm{A}\right|=-2$ and $\mathrm{B}=3$, then find the value of $\left|-5\mathrm{AB}\right|$.

Using properties of determinants, prove that

$\left|\begin{array}{ccc}1& 1+p& 1+p+q\\ 2& 3+2p& 4+3p+2q\\ 3& 6+3p& 10+6p+3p\end{array}\right|=1$.

Using properties of determinant, prove that

$\left|\begin{array}{ccc}pa& p^2+a^2& 1\\ pb& p^2+b^2& 1\\ pc& p^2+c^2& 1\end{array}\right|=p(a-b)(b-c)(c-a)$.

If $a, b$ and $c$ are real numbers and $∆=\left|\begin{array}{ccc}b+c& c+a& a+b\\ c+a& a+b& b+c\\ a+b& b+c& c+a\end{array}\right|=0,$ then show that either $a+b+c=0$ or $a=b=c.$

Solve the following for $x$:

$\left|\begin{array}{ccc}1& x& x^3\\ 1& a& a^3\\ 1& b& b^3\end{array}\right|=0, a\ne b$