← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q3a — Solution

15 marks · Section A

Question

Reduce the following matrix to echelon form: $A = \begin{bmatrix} 2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2 \end{bmatrix}$

Technique

Use Gaussian elimination (elementary row operations) to reach row-echelon form, then read off the rank.

Solution

$A=\begin{bmatrix} 2 & -2 & 2 & 1 \\ -3 & 6 & 0 & -1 \\ 1 & -7 & 10 & 2 \end{bmatrix}.$

Step 1 — Clear column 1 below the pivot $a_{11}=2$. Use fraction-free operations to keep integers.

$R_2 \to 2R_2 + 3R_1$: $2(-3,6,0,-1)+3(2,-2,2,1)=(0,6,6,1)$. $R_3 \to 2R_3 - R_1$: $2(1,-7,10,2)-(2,-2,2,1)=(0,-12,18,3)$.

$\sim\begin{bmatrix} 2 & -2 & 2 & 1 \\ 0 & 6 & 6 & 1 \\ 0 & -12 & 18 & 3 \end{bmatrix}.$

Step 2 — Clear column 2 below the pivot $6$.

$R_3 \to R_3 + 2R_2$: $(0,-12,18,3)+2(0,6,6,1)=(0,0,30,5)$.

$\sim\begin{bmatrix} 2 & -2 & 2 & 1 \\ 0 & 6 & 6 & 1 \\ 0 & 0 & 30 & 5 \end{bmatrix}.$

This is in row-echelon form: pivots are $2$ (col 1), $6$ (col 2), $30$ (col 3), each below-left of the next.

Step 3 — (Optional) tidy the last row. $R_3 \to \tfrac{1}{5}R_3 = (0,0,6,1)$, giving the equivalent echelon form $\begin{bmatrix} 2 & -2 & 2 & 1 \\ 0 & 6 & 6 & 1 \\ 0 & 0 & 6 & 1 \end{bmatrix}.$

There are 3 nonzero rows, so $\operatorname{rank}(A)=3$.

Answer

$\boxed{\;A \sim \begin{bmatrix} 2 & -2 & 2 & 1 \\ 0 & 6 & 6 & 1 \\ 0 & 0 & 30 & 5 \end{bmatrix},\qquad \operatorname{rank}(A)=3.\;}$ (Echelon forms are not unique; any form with three nonzero rows and pivots in columns 1, 2, 3 is acceptable.)