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

SAGAR SINGH - IIT DELHI

Last Activity: 14 Years ago

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

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

and

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

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

Algebra> Mtrices...

p is some matrix of order n and a is another matrix of the same order then what is1. pap'(p'=p transpose)2.pap-1(p-1=p inverse)

Vinay S R , 14 Years ago

SAGAR SINGH - IIT DELHI

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Algebra> Mtrices...

p is some matrix of order n and a is another matrix of the same order then what is1. pap'(p'=p transpose)2.pap-1(p-1=p inverse)

Vinay S R , 14 Years ago

SAGAR SINGH - IIT DELHI

Provide a better Answer & Earn Cool Goodies

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Ask a Doubt

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