UPSC 2023 Maths Optional Paper 1 Q4a — Step-by-Step Solution

15 marks · Section A

Rank of a matrix · Linear Algebra · asked 7× in 14 yrs · Read the full method →

Question

Find the rank of the matrix

A=\begin{bmatrix}1 & 2 & -1 & 0\\-1 & 3 & 0 & -4\\2 & 1 & 3 & -2\\1 & 1 & 1 & -1\end{bmatrix}

by reducing it to row-reduced echelon form.

Standard row reduction. Three pivots in non-zero rows → rank 3. The fourth row reduces to all zeros, indicating a linear dependence among the rows.

Step 1 — Eliminate column 1 below the pivot.

$R_2\to R_2+R_1$ : $(0,5,-1,-4)$ . $R_3\to R_3-2R_1$ : $(0,-3,5,-2)$ . $R_4\to R_4-R_1$ : $(0,-1,2,-1)$ .

\begin{bmatrix}1 & 2 & -1 & 0\\0 & 5 & -1 & -4\\0 & -3 & 5 & -2\\0 & -1 & 2 & -1\end{bmatrix}.

Step 2 — Eliminate column 2 below pivot in row 2.

$R_3\to R_3+(3/5)R_2$ : $\bigl(0,0,5-3/5,-2-12/5\bigr)=(0,0,22/5,-22/5)$ . $R_4\to R_4+(1/5)R_2$ : $\bigl(0,0,2-1/5,-1-4/5\bigr)=(0,0,9/5,-9/5)$ .

\begin{bmatrix}1 & 2 & -1 & 0\\0 & 5 & -1 & -4\\0 & 0 & 22/5 & -22/5\\0 & 0 & 9/5 & -9/5\end{bmatrix}.

Step 3 — Eliminate row 4’s column-3 entry.

$R_4\to R_4-(9/22)R_3$ : column 3 becomes $9/5-(9/22)(22/5)=9/5-9/5=0$ ; column 4 becomes $-9/5-(9/22)(-22/5)=-9/5+9/5=0$ .

\begin{bmatrix}1 & 2 & -1 & 0\\0 & 5 & -1 & -4\\0 & 0 & 22/5 & -22/5\\0 & 0 & 0 & 0\end{bmatrix}.

The matrix is now in row-echelon form. Three non-zero rows, so $\operatorname{rank}A=3$ .

\boxed{\;\operatorname{rank}A=3.\;}