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UPSC 2015 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Rank of a matrix · Linear Algebra · asked 7× in 13 yrs · Read the full method →

Question

Reduce the following matrix to row echelon form and hence find its rank:

[123421451557811417].\begin{bmatrix}1 & 2 & 3 & 4\\ 2 & 1 & 4 & 5\\ 1 & 5 & 5 & 7\\ 8 & 1 & 14 & 17\end{bmatrix}.

Technique

Gaussian elimination producing row echelon form; count non-zero rows.

Solution

Let AA denote the given matrix. Apply elementary row operations.

Step 1 — Eliminate column 1 below the pivot

R2R22R1,  R3R3R1,  R4R48R1R_2\to R_2-2R_1,\;R_3\to R_3-R_1,\;R_4\to R_4-8R_1:

(1234032303230151015).\begin{pmatrix}1 & 2 & 3 & 4\\ 0 & -3 & -2 & -3\\ 0 & 3 & 2 & 3\\ 0 & -15 & -10 & -15\end{pmatrix}.

Step 2 — Eliminate column 2 below the second pivot

Pivot in R2R_2 is 3-3. Use R3R3+R2R_3\to R_3+R_2 and R4R45R2R_4\to R_4-5R_2:

(1234032300000000).\begin{pmatrix}1 & 2 & 3 & 4\\ 0 & -3 & -2 & -3\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}.

This is row echelon form.

Step 3 — Read off rank

There are 2 non-zero rows. Hence

Answer

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