Find values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
Solution We have $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$
i.e. $\quad 3-x^{2}=3-8$
i.e. $x^{2}=8$
Hence $x=\pm 2 \sqrt{2}$
For the system of linear equations $a x+y+z=1$, $x+a y+z=1, x+y+a z=\beta$, which one of the following statements is NOT correct ?
If $1,\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\\{{\omega ^{2n}}}&1&{{\omega ^n}}\end{array}\,} \right|$ is equal to
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Let for any three distinct consecutive terms $a, b, c$ of an $A.P,$ the lines $a x+b y+c=0$ be concurrent at the point $\mathrm{P}$ and $\mathrm{Q}(\alpha, \beta)$ be a point such that the system of equations $ x+y+z=6, $ $ 2 x+5 y+\alpha z=\beta$ and $x+2 y+3 z=4$, has infinitely many solutions. Then $(P Q)^2$ is equal to________.
The value of the determinant$\left| {\,\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}\,} \right|$is equal to