Evaluate the determinants : $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$
$(ii)$ $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$
$=\left(x^{2}-x+1\right)(x+1)-(x-1)(x+1)$
$=x^{3}-x^{2}+x+x^{2}-x+1-\left(x^{2}-1\right)$
$=x^{3}+1-x^{2}+1$
$=x^{3}-x^{2}+2$
If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\m&n&p\\x&y&z\end{array}\,} \right| = k$, then $\left| {\,\begin{array}{*{20}{c}}{6a}&{2b}&{2c}\\{3m}&n&p\\{3x}&y&z\end{array}\,} \right| = $
If the system of equations $x - ky - z = 0$, $kx - y - z = 0$ and $x + y - z = 0$ has a non zero solution, then the possible value of k are
If the system of linear equations $x-2 y+z=-4 $ ; $2 x+\alpha y+3 z=5 $ ; $3 x-y+\beta z=3$ has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to
If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x - 2}\\{2{x^2} + 3x - 1}&{3x}&{3x - 3}\\{{x^2} + 2x + 3}&{2x - 1}&{2x - 1}\end{array}\,} \right| = Ax - 12$, then the value of $A $ is
If $ 5$ is one root of the equation $\left| {\,\begin{array}{*{20}{c}}x&3&7\\2&x&{ - 2}\\7&8&x\end{array}\,} \right| = 0$, then other two roots of the equation are