Let $ \ B \ $ be the matrix obtained from the matrix $ \ A \ $ by interchanging the $ \ i^{th} \ $ and $ \ j^{th} \ $ rows

Question

Let $ \ B \ $ be the matrix obtained from the matrix $ \ A \ $ by interchanging the $ \ i^{th} \ $ and $ \ j^{th} \ $ rows , then show that

$ B=MA \ $ , where $ \ M \ $ is the matrix obtained from the identity matrix $ \ I \ $ by interchanging $ \ i^{th} \ $ and $ \ j^{th} \ $ rows.

Answer:

Let $ \ A=[a_1 \ \ a_2 \ \ .... \ \ a_i \ \ ... .\ \ a_j .... \ a_n ]^T \ $ , where $ \ a_k , \ \ k=1,2,3,....,n \ $ are row vectors of $ \ A \ $

Then,

$ B=[a_1 \ \ a_2 \ \ ....a_j \ \ .... \ \ a_i \ \ .... \ a_n ]^T=[e_{ji}] [a_1 \ \ a_2 \ \ .... \ \ a_i \ \ ... .\ \ a_j .... \ a_n ]^T \ = MA$ ,

where $ e_{ij} =\delta_j^i \ $

Am I right?

Help me out

Answer 1

Let $$ A_1=\left( \begin{array}{ccccc} A_{11},&A_{12},&A_{13},& \cdots ,& A_{1n} \end{array}\right), \\ A_2=\left( \begin{array}{ccccc} A_{21},&A_{22},&A_{23},& \cdots ,& A_{2n} \end{array}\right), \\ A_3=\left( \begin{array}{ccccc} A_{31},&A_{32},&A_{33},& \cdots ,& A_{3n} \end{array}\right), \\ \vdots \\ \\ \vdots \\ A_n=\left( \begin{array}{ccccc} A_{n1},&A_{n2},&A_{n3},& \cdots ,& A_{nn} \end{array}\right), $$ the rows of matrix $$ A=\left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ A_{21}& A_{22}& A_{23}& \cdots & A_{2n}\\ A_{31}& A_{32}& A_{23}& \cdots & A_{3n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup = \left\lgroup \begin{array}{c} A_1\\ A_2\\ A_3\\ \vdots\\ A_n\\ \end{array}\right\rgroup $$

Let $$ e_1=\left( \begin{array}{ccccc} 1,&0,&0,& \cdots ,& 0 \end{array}\right), \\ e_2=\left( \begin{array}{ccccc} 0,&1,&0,& \cdots ,& 0 \end{array}\right), \\ e_3=\left( \begin{array}{ccccc} 0,&0,&1,& \cdots ,& 0 \end{array}\right), \\ \vdots \\ \\ \vdots \\ e_n=\left( \begin{array}{ccccc} 0,&0,&0,& \cdots ,& 1 \end{array}\right), $$ the rows of matrix identity $$ I=\left\lgroup \begin{array}{ccc c c} 1& 0& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ 0& 0& 0& \cdots & 1\\ \end{array}\right\rgroup = \left\lgroup \begin{array}{c} e_1\\ e_2\\ e_3\\ \vdots\\ e_n\\ \end{array}\right\rgroup $$ Note that $$ \left( \begin{array}{ccccc} 1,&0,&0,& \cdots ,& 0 \end{array}\right) \left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ A_{21}& A_{22}& A_{23}& \cdots & A_{2n}\\ A_{31}& A_{32}& A_{23}& \cdots & A_{3n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup = \left( \begin{array}{ccccc} A_{11},&A_{12},&A_{13},& \cdots ,& A_{1n} \end{array}\right) \\ \left( \begin{array}{ccccc} 0,&1,&0,& \cdots ,& 0 \end{array}\right) \left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ A_{21}& A_{22}& A_{23}& \cdots & A_{2n}\\ A_{31}& A_{32}& A_{23}& \cdots & A_{3n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup = \left( \begin{array}{ccccc} A_{21},&A_{22},&A_{23},& \cdots ,& A_{2n} \end{array}\right) \\ \left( \begin{array}{ccccc} 0,&0,&1,& \cdots ,& 0 \end{array}\right) \left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ A_{21}& A_{22}& A_{23}& \cdots & A_{2n}\\ A_{31}& A_{32}& A_{23}& \cdots & A_{3n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup = \left( \begin{array}{ccccc} A_{31},&A_{32},&A_{33},& \cdots ,& A_{3n} \end{array}\right) \\ \vdots \\ \vdots \\ \left( \begin{array}{ccccc} 0,&0,&0,& \cdots ,& 1 \end{array}\right) \left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ A_{21}& A_{22}& A_{23}& \cdots & A_{2n}\\ A_{31}& A_{32}& A_{23}& \cdots & A_{3n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup = \left( \begin{array}{ccccc} A_{n1},&A_{n2},&A_{n3},& \cdots ,& A_{nn} \end{array}\right) $$ We have for $j>i$ $$ \left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{j1}& A_{j2}& A_{j3}& \cdots & A_{jn}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{i1}& A_{i2}& A_{i3}& \cdots & A_{in}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup = \left\lgroup\begin{array}{c} e_1A\\ \vdots \\ e_jA \\ \vdots \\e_iA \\\vdots\\ e_nA \end{array}\right\rgroup = \left\lgroup\begin{array}{c} e_1\\ \vdots \\ e_j \\ \vdots \\e_i \\ \vdots \\e_n \end{array}\right\rgroup \left\lgroup \begin{array}{ccc c c} A_{11}& A_{12}& A_{13}& \cdots & A_{1n}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{i1}& A_{i2}& A_{i3}& \cdots & A_{in}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{j1}& A_{j2}& A_{j3}& \cdots & A_{jn}\\ \vdots& \vdots& \vdots& \ddots & \vdots\\ A_{n1}& A_{n2}& A_{n3}& \cdots & A_{nn}\\ \end{array}\right\rgroup $$

Answer 2

If you left multiply an $ n \times m$ matrix $A$ by a $n \times n$ matrix $M$, then each row of $M$ acts on the rows of $A$. So if the matrix $M$ has a row with all zeros but $1$ in the $i-th$ column, it just picks out the $i-th$ row of $A$. Hence a matrix that interchanges the $i-th$ and $j-th$ rows of $I_n$ (the $n \times n$ identity matrix) will simply interchange the corresponding rows of $A$.

The key is that multiplying $A$ on the left operates on its rows. (Multiplying on the right by some $m \times p$ matrix operates on the columns of $A$.)

Let $M^{(ij)}$ denote the matrix that one gets by interchanging the $i-th$ and $j-th$ rows of the identity matrix. These kinds of matrices are very useful in linear algebra, and they can be used to prove many properties of determinants.