Left| {begin{array}{*{20}{c}}{1 + {{sin }^2}θ }&{{{sin }^2}θ }&{{{sin }^2}θ }{{{cos }^2}?

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3 and 4 .Determinants and Matrices

hard

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to

$1/2$

$1$

$-1/2$

$-1$

Solution

(c) $\left| {\,\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$

Using ${C_1} \to {C_1} – {C_2},{C_2} \to {C_2} – {C_3}$

==> $\left| {\,\begin{array}{*{20}{c}}1&0&{{{\sin }^2}\theta }\\{ – 1}&1&{{{\cos }^2}\theta }\\0&{ – 1}&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$

==> $2\,(1 + 2\sin 4\theta ) = 0 \Rightarrow \sin 4\theta = \frac{{ – 1}}{2}$.

Standard 12

Mathematics

If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.

hard

If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}}$is

medium

If $\left| {\,\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}\,} \right| = k\,abc{(a + b + c)^3}$, then the value of $k$ is

hard

Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$

medium

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

hard

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to

Solution

Similar Questions

If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.

If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}}$is

If $\left| {\,\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}\,} \right| = k\,abc{(a + b + c)^3}$, then the value of $k$ is

Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$

Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$

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