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UPSC 2014 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

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

Question

Using elementary row or column operations, find the rank of the matrix

[01−3−10011310211−20].\begin{bmatrix}0 & 1 & -3 & -1\\ 0 & 0 & 1 & 1\\ 3 & 1 & 0 & 2\\ 1 & 1 & -2 & 0\end{bmatrix}.

Technique

Standard row reduction to echelon form.

Solution

Strategy. Reduce to row echelon form via elementary row operations; rank = number of non-zero rows.

Step 1 — Swap to get a leading 1 in row 1

R1↔R4R_1\leftrightarrow R_4:

[11−200011310201−3−1]\begin{bmatrix}1 & 1 & -2 & 0\\ 0 & 0 & 1 & 1\\ 3 & 1 & 0 & 2\\ 0 & 1 & -3 & -1\end{bmatrix}

Step 2 — Eliminate below pivot in column 1

R3→R3−3R1R_3\to R_3-3R_1:

[11−2000110−26201−3−1]\begin{bmatrix}1 & 1 & -2 & 0\\ 0 & 0 & 1 & 1\\ 0 & -2 & 6 & 2\\ 0 & 1 & -3 & -1\end{bmatrix}

Step 3 — Swap rows to find leading entry in column 2

R2↔R4R_2\leftrightarrow R_4:

[11−2001−3−10−2620011]\begin{bmatrix}1 & 1 & -2 & 0\\ 0 & 1 & -3 & -1\\ 0 & -2 & 6 & 2\\ 0 & 0 & 1 & 1\end{bmatrix}

Step 4 — Eliminate below pivot in column 2

R3→R3+2R2R_3\to R_3+2R_2:

[11−2001−3−100000011]\begin{bmatrix}1 & 1 & -2 & 0\\ 0 & 1 & -3 & -1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\end{bmatrix}

Step 5 — Swap zero row to bottom

R3↔R4R_3\leftrightarrow R_4:

[11−2001−3−100110000]\begin{bmatrix}1 & 1 & -2 & 0\\ 0 & 1 & -3 & -1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\end{bmatrix}

Row echelon form with three non-zero rows.

Answer

  rank⁡(A)=3.  \boxed{\;\operatorname{rank}(A)=3.\;}
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