An orthogonal matrix is

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3 and 4 .Determinants and Matrices

easy

An orthogonal matrix is

$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{2\sin \alpha }\\{ - 2\sin \alpha }&{\cos \alpha }\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$

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

Solution

(b) A square matrix is to be orthogonal matrix if $A'A = I = AA'$

$ \Rightarrow $ $A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ – \sin \alpha }&{\cos \alpha }\end{array}} \right]$, $A' = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ – \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$

$ \Rightarrow $ $AA' = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right],\,A'A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$

$\therefore $ $AA' = A'A = I$.

Standard 12

Mathematics

Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B =I$, where $I$ is identity matrix and $B$ is any square matrix of same order as of $A$. Matrix product $AB$ is equal to

normal

Let $A$ be a $2 \times 2$ matrix with real entries such that $A ^{\prime}=\alpha A + I$, where $\alpha \in R -\{-1,1\}$. If det $\left(A^2-A\right)=4$, then the sum of all possible values of $\alpha$ is equal to

hard

(JEE MAIN-2023)

Find the transpose of the following matrices : $\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$

easy

Let $X$ and $Y$ be two arbitrary, $3 \times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3 \times 3$, nonzero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?

$(A)$ $Y^3 Z^4-Z^4 Y^3$ $(B)$ $X ^{44}+ Y ^{44}$

$(C)$ $X ^4 Z ^3- Z ^3 X ^4$ $(B)$ $X ^{23}+ Y ^{23}$

medium

(IIT-2015)

Given $A$ and $C$ are involutary matrices and $B$ is a non-singular matrix, then $(AB^{-1}C)^{-1}$ is equal to –

normal

An orthogonal matrix is

Solution

Similar Questions

Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B =I$, where $I$ is identity matrix and $B$ is any square matrix of same order as of $A$. Matrix product $AB$ is equal to

Let $A$ be a $2 \times 2$ matrix with real entries such that $A ^{\prime}=\alpha A + I$, where $\alpha \in R -\{-1,1\}$. If det $\left(A^2-A\right)=4$, then the sum of all possible values of $\alpha$ is equal to

Find the transpose of the following matrices : $\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$

Given $A$ and $C$ are involutary matrices and $B$ is a non-singular matrix, then $(AB^{-1}C)^{-1}$ is equal to –

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