The determinant $\left|\begin{array}{ccc}a& b& a\alpha +b\\ b& c& b\alpha +c\\ a\alpha +b& b\alpha +c& 0\end{array}\right|$ is equal to zero if

If $\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]{\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}p& -q\\ q& p\end{array}\right]$, find the values of $p$ and $q$.

Let $\mathrm{A}=\mathrm{diagonal}[a_1, a_2, a_3], a_i\ne 0$, then Prove that ${\mathrm{A}}^{-1}=\mathrm{diagonal} [{a_1}^{-1}, {a_2}^{-1}, {a_3}^{-1}]$.

Obtain the inverse of the matrices

$\left[\begin{array}{ccc}1& p& 0\\ 0& 1& p\\ 0& 0& 1\end{array}\right]$ and $\left[\begin{array}{ccc}1& 0& 0\\ q& 1& 0\\ 0& q& 1\end{array}\right]$

And hence, find the inverse of the matrix $\left[\begin{array}{ccc}1+pq& p& 0\\ q& 1+pq& p\\ 0& q& 1\end{array}\right]$.

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

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