If A = left[ {begin{array}{*{20}{c}}{cos α }&{ - sin α }{sin α }&{cos α }end{array}} rig

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3 and 4 .Determinants and Matrices

medium

If $A = \left[ {\begin{array}{{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$and $B = \left[ {\begin{array}{{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, then the correct relation is

${A^2} = {B^2}$

$A + B = B - A$

$AB = BA$

None of these

Solution

(c) Clearly, $AB = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ – \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ – \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{\cos (\alpha + \beta )}&{ – \sin (\alpha + \beta )}\\{\sin (\alpha + \beta )}&{\cos (\alpha + \beta )}\end{array}} \right] = BA$ (verify).

Standard 12

Mathematics

In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write :

$(i)$ The order of the matrix

$(ii)$ The number of elements,

$(iii)$ Write the elements $a_{13}, \,a_{21}, \,a_{33}, \,a_{24},\, a_{23}$

easy

If $A = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 4}&2\end{array}} \right],$then ${A^2} – 5A = $

easy

If $A = \left[ {\begin{array}{{20}{c}}1&0\\2&0\end{array}} \right],B = \left[ {\begin{array}{{20}{c}}0&0\\1&{12}\end{array}} \right]$, then

easy

If $A\, = \,\left[ {\begin{array}{{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]$ and $B\, = \,\left[ {\begin{array}{{20}{c}}
y\\
x\\
1
\end{array}} \right]$ be such that $AB\, = \,\left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right],$ then

hard

(JEE MAIN-2014)

If $A = \left[ {\begin{array}{{20}{c}}0&2&0\\0&0&3\\{ – 2}&2&0\end{array}} \right]$and $B = \left[ {\begin{array}{{20}{c}}1&2&3\\3&4&5\\5&{ – 4}&0\end{array}} \right]$, then the element of $3^{rd}$ row and third column in $AB$ will be

easy

If $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, then the correct relation is

Solution

Similar Questions

In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write : $(i)$ The order of the matrix $(ii)$ The number of elements, $(iii)$ Write the elements $a_{13}, \,a_{21}, \,a_{33}, \,a_{24},\, a_{23}$

If $A = \left[ {\begin{array}{*{20}{c}}3&{ – 5}\\{ – 4}&2\end{array}} \right],$then ${A^2} – 5A = $

If $A = \left[ {\begin{array}{*{20}{c}}1&0\\2&0\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}0&0\\1&{12}\end{array}} \right]$, then

If $A\, = \,\left[ {\begin{array}{*{20}{c}} 1&2&x\\ 3&{ – 1}&2 \end{array}} \right]$ and $B\, = \,\left[ {\begin{array}{*{20}{c}} y\\ x\\ 1 \end{array}} \right]$ be such that $AB\, = \,\left[ {\begin{array}{*{20}{c}} 6\\ 8 \end{array}} \right],$ then

If $A = \left[ {\begin{array}{*{20}{c}}0&2&0\\0&0&3\\{ – 2}&2&0\end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}}1&2&3\\3&4&5\\5&{ – 4}&0\end{array}} \right]$, then the element of $3^{rd}$ row and third column in $AB$ will be

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If $A = \left[ {\begin{array}{{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$and $B = \left[ {\begin{array}{{20}{c}}{\cos \beta }&{ - \sin \beta }\\{\sin \beta }&{\cos \beta }\end{array}} \right]$, then the correct relation is

In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write :

$(i)$ The order of the matrix

$(ii)$ The number of elements,

$(iii)$ Write the elements $a_{13}, \,a_{21}, \,a_{33}, \,a_{24},\, a_{23}$

If $A = \left[ {\begin{array}{{20}{c}}1&0\\2&0\end{array}} \right],B = \left[ {\begin{array}{{20}{c}}0&0\\1&{12}\end{array}} \right]$, then

If $A\, = \,\left[ {\begin{array}{{20}{c}}
1&2&x\\
3&{ – 1}&2
\end{array}} \right]$ and $B\, = \,\left[ {\begin{array}{{20}{c}}
y\\
x\\
1
\end{array}} \right]$ be such that $AB\, = \,\left[ {\begin{array}{*{20}{c}}
6\\
8
\end{array}} \right],$ then

If $A = \left[ {\begin{array}{{20}{c}}0&2&0\\0&0&3\\{ – 2}&2&0\end{array}} \right]$and $B = \left[ {\begin{array}{{20}{c}}1&2&3\\3&4&5\\5&{ – 4}&0\end{array}} \right]$, then the element of $3^{rd}$ row and third column in $AB$ will be