Find values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
Solution We have $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
i.e. $\quad 3-x^{2}=3-8$
i.e. $x^{2}=8$
Hence $x=\pm 2 \sqrt{2}$
Show that points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear
Let $\mathrm{A}$ be a square matrix of order $3 \times 3$ , then $|\mathrm{k A}|$ is equal to
If $a > 0$and discriminant of $a{x^2} + 2bx + c$is negative, then $\left| {\,\begin{array}{*{20}{c}}a&b&{ax + b}\\b&c&{bx + c}\\{ax + b}&{bx + c}&0\end{array}\,} \right|$ is
$\left| {\,\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}\,} \right| = $
For all values of $A,B,C$ and $P,Q,R$, the value of $\left| {\,\begin{array}{*{20}{c}}{\cos (A - P)}&{\cos (A - Q)}&{\cos (A - R)}\\{\cos (B - P)}&{\cos (B - Q)}&{\cos (B - R)}\\{\cos (C - P)}&{\cos (C - Q)}&{\cos (C - R)}\end{array}\,} \right|$ is