GATE | GATE CS 1997 | Question 27

Let A= (aij) be an n-rowed square matrix and I12 be the matrix obtained by interchanging the first and second rows of the n-rowed Identify matrix. Then AI12 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


Answer: (C)

Explanation:  A =  \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\  a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\  \hdotsfor{5} \\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \\ \end{bmatrix} I_{12} =  \begin{bmatrix} 0 & 1 & 0 & \dots & 0 \\  1 & 0 & 0 & \dots & 0 \\  \hdotsfor{5} \\ 0 & 0 & 0 & \dots & 1 \\  \end{bmatrix} \\\\
When the above matrices are multiplied, the result is A_{12}, which is the matrix A with it’s first and second row interchanged.
This is because the first row of I_{12} has a 1 in the second column and it’s second row has a 1 in the first column. So when the matrices are multiplied, the first and second rows get exchanged.

This explanation is provided by Chirag Manwani.

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