Left| {,begin{array}{*{20}{c}}1&1&1{cos (nx)}&{cos (n + 1)x}&{cos (n + 2)x}{sin (nx)

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3 and 4 .Determinants and Matrices

medium

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$ is not depend

On $x$

On $n$

Both on $x$ and $n$

None of these

Solution

(b ) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&1&1\\{\cos nx}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin nx}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$

Applying ${C_1} \to {C_1} + {C_3} – (2\cos x){C_2}$

$\Delta = \left| {\,\begin{array}{*{20}{c}}{2(1 – \cos x)}&1&1\\0&{\cos (n + 1)x}&{\cos (n + 2)x}\\0&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$

$\Delta = 2(1 – \cos x)[\cos (n + 1)x\sin (n + 2)x$

$ – \cos (n + 2)x\sin (n + 1)x]$

$\Delta = 2(1 – \cos x)\,[\sin (n + 2 – n – 1)x]$ $ = 2\sin x(1 – \cos x)$

i.e., $\Delta $ is independent of $n$.

Standard 12

Mathematics

Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

hard

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

medium

If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ – x}}} \right)}^2}}&{{{\left( {{a^x}\, – \,\,{a^{ – x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ – y}}} \right)}^2}}&{{{\left( {{b^y}\, – \,\,{b^{ – y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ – z}}} \right)}^2}}&{{{\left( {{c^z}\, – \,\,{c^{ – z}}} \right)}^2}}&1\end{array}} \right|$ $=$

normal

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

hard

By using properties of determinants, show that:

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

medium

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\cos (nx)}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin (nx)}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}\,} \right|$ is not depend

Solution

Similar Questions

Using properties of determinants, prove that: $\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

By using properties of determinants, show that: $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

Using properties of determinants, prove that: $\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

By using properties of determinants, show that: $\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

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Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

By using properties of determinants, show that:

$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

By using properties of determinants, show that:

$\left|\begin{array}{ccc}-a^{2} & a b & a c \\ b a & -b^{2} & b c \\ c a & c b & -c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$