Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$
$\Delta=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$
$=x\left(x^{2}-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)$
$=x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta$
$=x^{3}-x+x\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$=x^{3}-x+x$
$\left.=x^{3} \quad \text { (Independent of } \theta\right)$
Hence, $\Delta$ is independent of $\theta$
If $\left| \begin{array}{*{20}{c}}
{ - 2a}&{a + b}&{a + c}\\
{b + a}&{ - 2b}&{b + c}\\
{c + a}&{b + c}&{ - 2c}
\end{array}\right|$ $ = \alpha \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ne 0$ then $\alpha $ is equal to
For the system of linear equations
$2 x-y+3 z=5$
$3 x+2 y-z=7$
$4 x+5 y+\alpha z=\beta$
Which of the following is NOT correct ?
Number of triplets of $a, b \, \& \,c$ for which the system of equations,$ax - by = 2a - b$ and $(c + 1) x + cy = 10 - a + 3 b$ has infinitely many solutions and $x = 1, y = 3$ is one of the solutions, is :
If $|A|$ denotes the value of the determinant of the square matrix $A$ of order $3$ , then $ |-2A|=$
A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 - x}&{ - 6}&3\\{ - 6}&{3 - x}&3\\3&3&{ - 6 - x}\end{array}\,} \right| = 0$ is