Let θ=frac{π}{5} and A=left[begin{array}{cc}cos θ & sin θ -sin θ & cos θend{array}righ

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

hard

Let $\theta=\frac{\pi}{5}$ and $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \cdot$ If $B=A + A ^{4},$ then $\operatorname{det}( B )$

is one

lies in $(1,2)$

is zero

lies in $(2,3)$

(JEE MAIN-2020)

Solution

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

$A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

$B=A+A^{4}$

$=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\left[\begin{array}{cc}\cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta\end{array}\right]$

$B=\left[\begin{array}{cc}(\cos \theta+\cos 4 \theta) & (\sin \theta+\sin 4 \theta) \\ -(\sin \theta+\sin 4 \theta) & (\cos \theta+\cos 4 \theta)\end{array}\right]$

$|B|=(\cos \theta+\cos 4 \theta)^{2}+(\sin \theta+\sin 4 \theta)^{2}$

$|B|=2+2 \cos 3 \theta$

when $\theta=\frac{\pi}{5}$

$| B |=2+2 \cos \frac{3 \pi}{5}=2(1-\sin 18)$

$| B |=2\left(1-\frac{\sqrt{5}-1}{4}\right)=2\left(\frac{5-\sqrt{5}}{4}\right)=\frac{5-\sqrt{5}}{2}$

Standard 12

Mathematics

If $A = \left[ {\begin{array}{{20}{c}}i&0\\0&i\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}0&{ – i}\\{ – i}&0\end{array}} \right]$, then $(A + B)(A – B)$ is equal to

easy

Construct a $3 \times 2$ matrix whose elements are given by $a_{i j}=\frac{1}{2}|i-3 j|$.

medium

If $A = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\2&3&0&0\\4&5&6&0\\7&8&9&{10}\end{array}} \right]$, then $A$ is

normal

If $A =$ $\left[ {\,\begin{array}{{20}{c}}{{{\cos }^2}\alpha }&{\sin \alpha \,\,\cos \alpha }\\{\sin \alpha \,\,\cos \alpha }&{{{\sin }^2}\alpha }\end{array}\,} \right]$; $B =$ $\left[ {\,\begin{array}{{20}{c}}{{{\cos }^2}\beta }&{\sin \beta \,\,\cos \beta }\\{\sin \beta \,\,\cos \beta }&{{{\sin }^2}\beta }\end{array}\,} \right]$ are such that, $AB$ is a null matrix, then which of the following should necessarily be an odd integral multiple of $\frac{\pi }{2}$.

normal

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ – 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} – 2{x^2} – \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then

normal

Let $\theta=\frac{\pi}{5}$ and $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \cdot$ If $B=A + A ^{4},$ then $\operatorname{det}( B )$

Solution

Similar Questions

If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&i\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}0&{ – i}\\{ – i}&0\end{array}} \right]$, then $(A + B)(A – B)$ is equal to

Construct a $3 \times 2$ matrix whose elements are given by $a_{i j}=\frac{1}{2}|i-3 j|$.

If $A = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\2&3&0&0\\4&5&6&0\\7&8&9&{10}\end{array}} \right]$, then $A$ is

Let $A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&2&{ – 1}\\ 3&0&k \end{array}} \right]$ and $f(x) = {x^3} – 2{x^2} – \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then

Start a Free Trial Now

If $A = \left[ {\begin{array}{{20}{c}}i&0\\0&i\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}0&{ – i}\\{ – i}&0\end{array}} \right]$, then $(A + B)(A – B)$ is equal to

Let $A = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
2&2&{ – 1}\\
3&0&k
\end{array}} \right]$ and $f(x) = {x^3} – 2{x^2} – \alpha x + \beta = 0$ . If $A$ satisfies $f(x)=0$ ,then