3,644 views

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

Typo here "Identity" matrix and not "Identify"

(C)
what is the meaning of I12 matrix

$C$ is the answer.

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

### 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.

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.
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.

1
2,736 views
2
17,243 views
3
5,431 views