The value of the determinant$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 - x}&1\\1&1&{1 + y}\end{array}\,} \right|$is
$3 - x + y$
$(1 - x)(1 + y)$
$xy$
$ - xy$
Consider system of equations in $x$ , $y$ and $z$
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
Evaluate the determinants : $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$
Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha\end{array}\right]$ and $|2 A|^3=2^{21}$ where $\alpha, \beta \in Z$, Then a value of $\alpha $ is
Let $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right],$ where $0 \leq \theta \leq 2 \pi$. Then