UPSC NDA I 2015 Mathematics Paper

Concept:An elementary matrix is obtained by performing a single elementary row operation on an identity matrix.It must have all diagonal entries equal to $1$, except possibly for a scaling row operation under certain conditions.Explanation:Check each option to see if it can be transformed back to the identity matrix by a single elementary row operation.Option A: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ has a zero on the diagonal (row 2, column 2).This cannot come from an identity matrix by a single elementary operation because a zero diagonal entry would require a row scaling by $0$, which is not allowed.Option B: $\begin{bmatrix} 1 & 5 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.Apply the row operation $R_1 \rightarrow R_1 - 5R_2$.It becomes $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3$.Thus, it differs from $I_3$ by exactly one elementary row operation.Option C: $\begin{bmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.The first two rows are swapped but also scaled (the $\,2\,$ in row 1, column 2).A single row swap would give $0$ and $1$ off‑diagonal entries but no scaling; this requires at least two operations.Option D: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 5 & 2 \end{bmatrix}$.The last row has a $5$ and a $2$ instead of a $1$ on the diagonal.To obtain identity would need more than one operation (e.g., scale row 3 by $1/2$, then subtract $5$ times row 2).Only option B satisfies the definition of an elementary matrix.Answer:Option B is the elementary matrix.