Typo here **"Identity"** matrix and not **"Identify"**

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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

- Row is the same as its second row
- Row is the same as the second row of $A$
- Column is the same as the second column of $A$
- Row is all zero

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.

$$A=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix},I_{12}=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},AI_{12}=\begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$$

$AI_{12}$ matrix will result into the same matrix as $A$ but first column will be exchanged with second column.

$$A=\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix},I_{12}=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},AI_{12}=\begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$$

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.

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.

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.

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

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 $B$

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