GATE CSE 1997 | Question: 4.2

Question

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

• A. Row is the same as its second row
• B. Row is the same as the second row of $A$
• C. Column is the same as the second column of $A$
• D. Row is all zero

Comments

what is the meaning of I12 matrix

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Interchanging 1st row and 2nd row , R1<->R2

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Ans

Answer 1

$C$ is the answer.

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

Comments

When a row of a matrix is mul. with nth column of identity matrix we get its nth element of the row.

Answer 2

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.

Comments

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.

Answer 3

This question is basically an application of a permutation matrix 
https://mathworld.wolfram.com/PermutationMatrix.html
U can take a small matrix and check in exam hall just to get the answer

Answer 4

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

Comments

Yeah! 3blue1brown really does a great job.

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Nice explanation !!

Answer 5

A permutation matrix, denoted as P, is a square matrix that represents a permutation of the rows or columns of an identity matrix.

When you multiply a permutation matrix on the left side of a matrix A (PA), it actually permutes the rows of matrix A according to the permutation represented by the permutation matrix P.

PA = A (with row of A swapped acc to permutation of P)

Similarly, when you multiply a permutation matrix on the right side of a matrix A (AP), it permutes the columns of matrix A according to the same permutation represented by the permutation matrix P.

AP = A (with columns of A swapped acc to permutation of P)

Here, $AI_{12}$   hence multiplying permutation matrix on right side of A and 1st row and 2nd rows of P are exchanged (exchanging 1st 2nd rows or exchanging 1st 2nd column of $I_{12}$ is equivalent).

Hence $AI_{12}$ will record exchanging 1st and 2nd columns of matrix A.

Answer 6

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.