If $P = left[ {begin{array}{*{20}{c}} {frac{{√ 3 }}{2}}&{frac{1}{2}} {

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3 and 4 .Determinants and Matrices

medium

If $P = \left[ {\begin{array}{{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{ - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right],\,A = \,\left[ {\begin{array}{{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $Q=PAP^T,$ then $P^T$ $Q^{2015}$ $P$ is

$\,\left[ {\begin{array}{*{20}{c}}
0&{2015}\\
0&0
\end{array}} \right]$

$\,\left[ {\begin{array}{*{20}{c}}
{2015}&0\\
1&{2015}
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
1&{2015}\\
0&1
\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}
{2015}&1\\
0&{2015}
\end{array}} \right]$

(JEE MAIN-2016)

Solution

$P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

$P^{T}=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$

$P P^{T}=P^{T} P=I$

$\mathrm{Q}^{2015}=\left(P A P^{T}\right)\left(P A P^{T}\right)-(2015 \text { terms })$

$=P A^{2015} P^{T}$

$P^{T} \mathrm{Q}^{2015} P=A^{2015}$

$A^{2}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$

$A^{3}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$

$A^{2015}=\left[\begin{array}{cc}1 & 2015 \\ 0 & 1\end{array}\right]$

Standard 12

Mathematics

If $A = \left[ {\begin{array}{{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{\frac{{ – 1}}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right]$ , then $(BB^TA)^5$ is equal to

normal

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

normal

For any square matrix $A$, $A{A^T}$ is a

normal

Let $A , B , C$ be $3 \times 3$ matrices such that $A$ is symmetric and $B$ and $C$ are skew-symmetric.Consider the statements

$(S1): A ^{13} B ^{26}- B ^{26} A ^{13}$ is symmetric

$(S2):A ^{26} C ^{13}- C ^{13} A ^{26}$ is symmetric

Then,

hard

(JEE MAIN-2023)

Which of the following is incorrect

medium

If $P = \left[ {\begin{array}{*{20}{c}} {\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\ { - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}} \end{array}} \right],\,A = \,\left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right]$ and $Q=PAP^T,$ then $P^T$ $Q^{2015}$ $P$ is

Solution

Similar Questions

If $A = \left[ {\begin{array}{*{20}{c}} 1&1\\ 0&1 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} {\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\ {\frac{{ – 1}}{2}}&{\frac{{\sqrt 3 }}{2}} \end{array}} \right]$ , then $(BB^TA)^5$ is equal to

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For any square matrix $A$, $A{A^T}$ is a

Let $A , B , C$ be $3 \times 3$ matrices such that $A$ is symmetric and $B$ and $C$ are skew-symmetric.Consider the statements $(S1): A ^{13} B ^{26}- B ^{26} A ^{13}$ is symmetric $(S2):A ^{26} C ^{13}- C ^{13} A ^{26}$ is symmetric Then,

Which of the following is incorrect

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If $P = \left[ {\begin{array}{{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{ - \frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right],\,A = \,\left[ {\begin{array}{{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $Q=PAP^T,$ then $P^T$ $Q^{2015}$ $P$ is

If $A = \left[ {\begin{array}{{20}{c}}
1&1\\
0&1
\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}
{\frac{{\sqrt 3 }}{2}}&{\frac{1}{2}}\\
{\frac{{ – 1}}{2}}&{\frac{{\sqrt 3 }}{2}}
\end{array}} \right]$ , then $(BB^TA)^5$ is equal to

Let $A , B , C$ be $3 \times 3$ matrices such that $A$ is symmetric and $B$ and $C$ are skew-symmetric.Consider the statements

$(S1): A ^{13} B ^{26}- B ^{26} A ^{13}$ is symmetric

$(S2):A ^{26} C ^{13}- C ^{13} A ^{26}$ is symmetric

Then,