If A = left[ {begin{array}{*{20}{c}}1&{tan θ /2}{ - tan θ /2}&1end{array}} right] and AB =

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3 and 4 .Determinants and Matrices

easy

If $A = \left[ {\begin{array}{*{20}{c}}1&{\tan \theta /2}\\{ - \tan \theta /2}&1\end{array}} \right]$ and $AB = I$, then $B = $

${\cos ^2}\frac{\theta }{2}.A$

${\cos ^2}\frac{\theta }{2}.{A^T}$

${\cos ^2}\frac{\theta }{2}.I$

None of these

Solution

(b) $|A| = 1 + {\tan ^2}\frac{\theta }{2} = {\sec ^2}\frac{\theta }{2}$

$AB = I$ $ \Rightarrow $ $B = I{A^{ – 1}}$

$\frac{{\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}1&{ – \tan \frac{\theta }{2}}\\{\tan \frac{\theta }{2}}&1\end{array}} \right]}}{{{{\sec }^2}\frac{\theta }{2}}}$= ${\cos ^2}\frac{\theta }{2}\,.\,{A^T}$.

Standard 12

Mathematics

The matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}} \right]$is non singular, if

easy

Find $A B,$ if $A=\left[\begin{array}{ll}6 & 9 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{lll}2 & 6 & 0 \\ 7 & 9 & 8\end{array}\right]$

easy

Show that

$\left[ {\begin{array}{{20}{c}}
5&{ – 1} \\
6&7
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]$ $ \ne \left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
5&{ – 1} \\
6&7
\end{array}} \right]$

medium

$\cos \theta \left[ {\begin{array}{{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

easy

The number of $3 \times 3$ matrices $A$, whose entries are either $1$ or $-1$ and for which the system $A\left[ {\begin{array}{{20}{c}}
x\\
y\\
z
\end{array}} \right] = \left[ {\begin{array}{{20}{c}}
1\\
{ – 1}\\
0
\end{array}} \right]$ has exactly three distinct solutions, is

normal

If $A = \left[ {\begin{array}{*{20}{c}}1&{\tan \theta /2}\\{ - \tan \theta /2}&1\end{array}} \right]$ and $AB = I$, then $B = $

Solution

Similar Questions

The matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}} \right]$is non singular, if

Find $A B,$ if $A=\left[\begin{array}{ll}6 & 9 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{lll}2 & 6 & 0 \\ 7 & 9 & 8\end{array}\right]$

Show that $\left[ {\begin{array}{*{20}{c}} 5&{ – 1} \\ 6&7 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 2&1 \\ 3&4 \end{array}} \right]$ $ \ne \left[ {\begin{array}{*{20}{l}} 2&1 \\ 3&4 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 5&{ – 1} \\ 6&7 \end{array}} \right]$

$\cos \theta \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

The number of $3 \times 3$ matrices $A$, whose entries are either $1$ or $-1$ and for which the system $A\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ { – 1}\\ 0 \end{array}} \right]$ has exactly three distinct solutions, is

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Show that

$\left[ {\begin{array}{{20}{c}}
5&{ – 1} \\
6&7
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]$ $ \ne \left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
5&{ – 1} \\
6&7
\end{array}} \right]$

$\cos \theta \left[ {\begin{array}{{20}{c}}{\cos \theta }&{\sin \theta }\\{ – \sin \theta }&{\cos \theta }\end{array}} \right] + \sin \theta \left[ {\begin{array}{{20}{c}}{\sin \theta }&{ – \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right] = $

The number of $3 \times 3$ matrices $A$, whose entries are either $1$ or $-1$ and for which the system $A\left[ {\begin{array}{{20}{c}}
x\\
y\\
z
\end{array}} \right] = \left[ {\begin{array}{{20}{c}}
1\\
{ – 1}\\
0
\end{array}} \right]$ has exactly three distinct solutions, is