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Vinay S R Grade:

p is some matrix of order n and a is another matrix of the same order then what is

1. pap'(p'=p transpose)

2.pap-1(p-1=p inverse)

7 years ago

Answers : (1)

879 Points

Dear vinay,

The transpose of a matrix ${\bf A}$, denoted by ${\bf A}^T$, is obtained by switching the positions of elements $a_{ij}$ and $a_{ji}$ for all $i,j \in \{1,\cdots,n\}$. In other words, the ith column of ${\bf A}$ becomes the ith row of ${\bf A}^T$, or equivalently, the ith row of ${\bf A}$ becomes the ith column of ${\bf A}^T$:

\begin{displaymath}{\bf A}^T=[{\bf a}_1 \cdots {\bf a}_n]^T= \left[ \begin{array}{c} {\bf a}_1^T  ..  .. {\bf a}_n^T \end{array} \right] \end{displaymath}

where vector ${\bf a}_i$ is the ith column of ${\bf A}$ and its transpose ${\bf a}_i^T$ is the ith row of ${\bf A}^T$.

For any two matrices ${\bf A}$ and ${\bf B}$, we have

\begin{displaymath}({\bf A}{\bf B})^T={\bf B}^T {\bf A}^T \end{displaymath}

If ${\bf AB}={\bf BA}={\bf I}$, where ${\bf I}$ is an identity matrix:

\begin{displaymath}{\bf I}=diag[1,\cdots,1]=\left[ \begin{array}{cccc} 1 & 0 &... ...\cdot & \cdot & \cdot  0 & 0 & \cdot & 1 \end{array} \right] \end{displaymath}

then ${\bf B}={\bf A}^{-1}$ is the inverse of ${\bf A}$. ${\bf A}^{-1}$ exists iff $ det({\bf A}) \neq 0 $, i.e., $ rank({\bf A}) = n $.

For any two matrices ${\bf A}$ and ${\bf B}$, we have

\begin{displaymath}({\bf AB})^{-1}={\bf B}^{-1}{\bf A}^{-1} \end{displaymath}


\begin{displaymath}({\bf A}^{-1})^T=({\bf A}^T)^{-1} \end{displaymath}

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi

7 years ago
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