in Linear Algebra edited by
3,644 views
22 votes
22 votes

Let $A=(a_{ij})$ be an $n$-rowed square matrix and $I_{12}$ be the matrix obtained by interchanging the first and second rows of the $n$-rowed Identity matrix. Then $AI_{12}$ is such that its first

  1. Row is the same as its second row
  2. Row is the same as the second row of $A$
  3. Column is the same as the second column of $A$
  4. Row is all zero
in Linear Algebra edited by
3.6k views

3 Comments

Typo here "Identity" matrix and not "Identify"

2
2
(C)
0
0
what is the meaning of I12 matrix
0
0

4 Answers

31 votes
31 votes
Best answer
$C$ is the answer.

$AI_{12}$ matrix will result into the same matrix as $A$ but first column will be exchanged with second column.
$$A=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix},I_{12}=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},AI_{12}=\begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$$
edited by
2 votes
2 votes
We can solve it by taking an example.

Let $n=2$ and matrix $A$ be $\begin{bmatrix} 2 &1 \\ 1&3 \end{bmatrix}$

$\implies I_{12}=\begin{bmatrix} 0 &1 \\ 1&0 \end{bmatrix}$... (obtained from $I_2 = \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$)

 

$\implies AI_{12} = $ $\begin{bmatrix} 2 &1 \\ 1&3 \end{bmatrix}*$$\begin{bmatrix} 0 &1 \\ 1&0 \end{bmatrix}=$$\begin{bmatrix} 1 &2 \\ 3&1\end{bmatrix}$

 

By seeing the matrix we can clearly say that Option $C.$ is correct answer.

1 comment

we can solve it without any example or without any multiplication of matrices.

Here, $I_{12}$ is working like permutation matrix. Interchanging 1st row and 2nd row of I is same as interchanging 1st and 2nd column and since it is multiplying at right side of A, so it permute the 1st and 2nd columns of matrix A.
8
8
0 votes
0 votes

But note that $I_4$ is nothing but $n$x$1$ vectors stacked side by side 

$I_4 = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0& 1& 0& 0\\ 0& 0 &1 &0 \\ 0& 0 &0 &1 \end{bmatrix}$

Also note the behaviour on passing these $n$x$1$ vectors to an arbitrary matrix transformation $A=\begin{bmatrix} a & b& c & d\\ e & f & g & h\\ i & j & k & l \\ m&n &o &p \end{bmatrix}$

 

$Then \; A\begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} a & b& c & d\\ e & f & g & h\\ i & j & k & l \\ m&n &o &p \end{bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} a\\ e\\ i\\ m \end{bmatrix} $

 

$Similarly, A\begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix} = \begin{bmatrix} b\\ f\\ j\\ n \end{bmatrix} .. and \; so \; on$

So if we interchange first two rows of the $I$ matrix, we simply interchange the first two column vectors of $I$ and hence of A..

So the answer is $B$

For an intuitive understanding this playlist is highly recommended

by

1 comment

Yeah! 3blue1brown really does a great job.
0
0
0 votes
0 votes
Take a $2\times2$ matrix and multiply it with $\bigl(\begin{smallmatrix} 0 &1 \\ 1& 0 \end{smallmatrix}\bigr)$. You will get to see that the columns of the original matrix has interchanged. Hence option $(C)$ is correct.
Answer:

Related questions