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

$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a – 1)}^2}}&{{{(b – 1)}^2}}&{{{(c – 1)}^2}}\end{array}\,} \right| = $

medium

If $p, q, r, s$ are in $A.P.$ and $f (x) =$ $\left| {\,\begin{array}{*{20}{c}} {p\,\, + \,\,\sin \,x}&{q\,\, + \,\,\sin \,x}&{p\,\, – \,\,r\,\, + \,\,\sin \,x}\\ {q\,\, + \,\,\sin \,x}&{r\,\, + \,\,\sin \,x}&{ – \,1\,\, + \,\,\sin \,x}\\ {r\,\, + \,\,\sin \,x}&{s\,\, + \,\,\sin \,x}&{s\,\, – \,\,q\,\, + \,\,\sin \,x} \end{array}\,} \right|$ such that $f (x)dx = – 4$ then the common difference of the $A.P.$ can be :

normal

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|=0$

easy

$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $

hard

$\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}\,} \right| = $