$\left| {\,\begin{array}{*{20}{c}}{11}&{12}&{13}\\{12}&{13}&{14}\\{13}&{14}&{15}\end{array}\,} \right| = $
$1$
$0$
$-1$
$67$
(b) Apply ${C_3} \to {C_3} – {C_2}$ and ${C_2} \to {C_2} – {C_1}$.
સમીકરણની સંહતિ $\begin{array}{l}\alpha x + y + z = \alpha – 1\\x + \alpha y + z = \alpha – 1\\x + y + \alpha z = \alpha – 1\end{array}$ નો ઉકેલ ખાલીગણ હોય તો $\alpha $ કિમત મેળવો.
જો $0 \leq \theta \leq 2 \pi$ માટે $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$ હોય, તો
જો $\omega $ એકનું કાલ્પનિક ઘનમૂળ હોય , તો $\left| {\,\begin{array}{*{20}{c}}a&{b{\omega ^2}}&{a\omega }\\{b\omega }&c&{b{\omega ^2}}\\{c{\omega ^2}}&{a\omega }&c\end{array}\,} \right|$ મેળવો.
જો $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$ તો $x =$
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