Evaluate the determinants
$\left|\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right|$
Let $A=\left[\begin{array}{ccc}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right]$
By expanding along the first column, we have:
$|A|=2\left|\begin{array}{cc}2 & -1 \\ -5 & 0\end{array}\right|-0\left|\begin{array}{cc}-1 & -2 \\ -5 & 0\end{array}\right|+3\left|\begin{array}{cc}-1 & -2 \\ 2 & -1\end{array}\right|$
$=2(0-5)-0+3(1+4)$
$=-10+15=5$
If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right],$ find $|A|$.
The value of $a$ for which the system of equations
$a^3x + ( a + 1)^3y + (a + 2)^3z = 0$ ; $ax + (a + 1) y + ( a + 2) z = 0$ ; $x + y + z = 0$, has a non zero solution is
If the system of equations $\mathrm{x}+4 \mathrm{y}-\mathrm{z}=\lambda$, $7 x+9 y+\mu z=-3,5 x+y+2 z=-1$ has infinitely many solutions, then $(2 \mu .+3 \lambda)$ is equal to :
Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then
Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations
$2 x-y+2 z=2$
$x-2 y+\lambda z=-4$
$x+\lambda y+z=4$
has no solution. Then the set $S$