UPSC 2019 Maths Optional Paper 1 Q3c-i — Step-by-Step Solution

15 marks · Section A

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

Question

Let $A=\begin{pmatrix}5&7&2&1\\1&1&-8&1\\2&3&5&0\\3&4&-3&1\end{pmatrix}$ . Find the rank of matrix $A$ .

Gaussian elimination to row echelon form; rank $=$ number of nonzero rows.

Use $R_2$ (which has a leading $1$ ) as pivot. Reorder/eliminate the first column. Eliminate $x_1$ using $R_2$ :

R_1\to R_1-5R_2,\quad R_3\to R_3-2R_2,\quad R_4\to R_4-3R_2.

\begin{pmatrix}1&1&-8&1\\5&7&2&1\\2&3&5&0\\3&4&-3&1\end{pmatrix}\ \xrightarrow{\ }\ \begin{pmatrix}1&1&-8&1\\0&2&42&-4\\0&1&21&-2\\0&1&21&-2\end{pmatrix}.

(Here $R_1-5R_2=(0,2,42,-4)$ , $R_3-2R_2=(0,1,21,-2)$ , $R_4-3R_2=(0,1,21,-2)$ .)

Rows 2,3,4 are all proportional: $(0,2,42,-4)=2\,(0,1,21,-2)$ , and rows 3 and 4 are identical. Eliminate:

R_2\to R_2-2R_3,\quad R_4\to R_4-R_3:

\begin{pmatrix}1&1&-8&1\\0&1&21&-2\\0&0&0&0\\0&0&0&0\end{pmatrix}.

The row echelon form has exactly two nonzero rows (pivots in columns 1 and 2).

\boxed{\;\operatorname{rank}(A)=2.\;}