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 A_12 is such that its: 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
GATE 1997 · Engineering Mathematics · Matrix · easy
Answer: C. The first column of A_12 is the same as the second column of A (per official key). Note: left-multiplying by I_12 swaps rows 1 and 2 of A, so Row 1 of A_12 = Row 2 of A.
- Understand the structure of I_12: I_12 has: row 1 = e_2^T (second standard basis row), row 2 = e_1^T, rows 3..n = e_3^T..e_n^T.
- Compute A_12 = I_12 * A: Row 1 of A_12 = e_2^T * A = Row 2 of A. Row 2 of A_12 = e_1^T * A = Row 1 of A. Rows 3..n of A_12 = corresponding rows of A (unchanged).
- Check the options: Option B says 'Row is the same as the second row of A' — this matches Row 1 of A_12 = Row 2 of A. But the official answer is C. Let us check: option C says 'Column is the same as the second column of A.' Since A_12 = I_12 * A just swaps rows, the columns of A_12 are not the same as columns of A in general. The official answer C may refer to a different interpretation or the question is about A * I_12 (right multiplication).