If A = left[ {begin{array}{*{20}{c}}{{{cos }^2}θ }&{sin θ cos θ }{sin θ cos θ }&{{{sin

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3 and 4 .Determinants and Matrices

medium

If $A = \left[ {\begin{array}{{20}{c}}{{{\cos }^2}\theta }&{\sin \theta \cos \theta }\\{\sin \theta \cos \theta }&{{{\sin }^2}\theta }\end{array}} \right]$, $B = \left[ {\begin{array}{{20}{c}}{{{\cos }^2}\phi }&{\sin \phi \cos \phi }\\{\sin \phi \cos \phi }&{{{\sin }^2}\phi }\end{array}} \right]$ and $\theta $ and $\phi $ differs by $\frac{\pi }{2},$ then $AB = $

$I$

$O$

$-I$

None of these

Solution

(b) $AB = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\sin \theta \cos \theta }\\{\sin \theta \cos \theta }&{{{\sin }^2}\theta }\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\phi }&{\sin \phi \cos \phi }\\{\sin \phi \cos \phi }&{{{\sin }^2}\phi }\end{array}} \right]$
= $\left[ {\begin{array}{*{20}{c}}{\cos \theta \cos \phi \cos (\theta – \phi )}&{\cos \theta \sin \phi \cos (\theta – \phi )}\\{\cos \theta \sin \phi \cos (\theta – \phi )}&{\sin \theta \sin \phi \cos (\theta – \phi )}\end{array}} \right]$
= $\cos (\theta – \phi )\,\left[ {\begin{array}{*{20}{c}}{\cos \theta \cos \phi }&{\cos \theta \sin \phi }\\{\cos \theta \sin \phi }&{\sin \theta \sin \phi }\end{array}} \right]$
= $O$, $\left( {\because \theta – \phi = \frac{\pi }{2}} \right)$

Standard 12

Mathematics

The solution $(s)$ of the equation $\left| {\begin{array}{*{20}{c}}x&a&b\\ a&x&a\\b&b&x\end{array}} \right|$ $= 0$ is/are :

normal

Let $S=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; a, b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap \limits_{n=1}^{100} T_{n}$ is

hard

(JEE MAIN-2022)

If $A = \left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ – {a^2}}&{ – ab}\end{array}} \right]$ and ${A^n} = O$, then the minimum value of $n$ is

medium

If $A = \left[ {\begin{array}{{20}{c}}
\alpha &0\\
1&1
\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}
1&0\\
5&1
\end{array}} \right]$ , then the value of $\alpha $ for which $A^2 = B$ is

normal

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$

Find $BA$

easy

Solution

Similar Questions

The solution $(s)$ of the equation $\left| {\begin{array}{*{20}{c}}x&a&b\\ a&x&a\\b&b&x\end{array}} \right|$ $= 0$ is/are :

Let $S=\left\{\left(\begin{array}{cc}-1 & a \\ 0 & b\end{array}\right) ; a, b \in\{1,2,3, \ldots 100\}\right\}$ and let $T_{n}=\left\{A \in S: A^{n(n+1)}=I\right\}$. Then the number of elements in $\bigcap \limits_{n=1}^{100} T_{n}$ is

If $A = \left[ {\begin{array}{*{20}{c}}{ab}&{{b^2}}\\{ – {a^2}}&{ – ab}\end{array}} \right]$ and ${A^n} = O$, then the minimum value of $n$ is

If $A = \left[ {\begin{array}{*{20}{c}} \alpha &0\\ 1&1 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 1&0\\ 5&1 \end{array}} \right]$ , then the value of $\alpha $ for which $A^2 = B$ is

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$ Find $BA$

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If $A = \left[ {\begin{array}{{20}{c}}
\alpha &0\\
1&1
\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}
1&0\\
5&1
\end{array}} \right]$ , then the value of $\alpha $ for which $A^2 = B$ is

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$

Find $BA$