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UPSC 2021 Maths Optional Paper 1 Q4a-i — Step-by-Step Solution

10 marks · Section A

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

Question

A=[132410022026262391106]A=\begin{bmatrix}1 & 3 & 2 & 4 & 1\\ 0 & 0 & 2 & 2 & 0\\ 2 & 6 & 2 & 6 & 2\\ 3 & 9 & 1 & 10 & 6\end{bmatrix}. Reduce to RREF; find rank.

Technique

Standard Gaussian elimination to RREF; track pivot positions; count non-zero rows.

Solution

Step 1 — Use R1R_1 to clear column 1

R3R32R1R_3\to R_3-2R_1: (0,0,2,2,0)(0,0,-2,-2,0).

R4R43R1R_4\to R_4-3R_1: (0,0,5,2,3)(0,0,-5,-2,3).

[13241002200022000523].\begin{bmatrix}1 & 3 & 2 & 4 & 1\\ 0 & 0 & 2 & 2 & 0\\ 0 & 0 & -2 & -2 & 0\\ 0 & 0 & -5 & -2 & 3\end{bmatrix}.

Step 2 — Use R2R_2 to clear column 3

R2R_2 has pivot in column 3 (value 2). Divide R2R_2 by 2: (0,0,1,1,0)(0,0,1,1,0).

R3R3+2R2R_3\to R_3+2R_2: (0,0,0,0,0)(0,0,0,0,0).

R4R4+5R2R_4\to R_4+5R_2: (0,0,0,3,3)(0,0,0,3,3).

R1R12R2R_1\to R_1-2R_2: (1,3,0,2,1)(1,3,0,2,1).

[13021001100000000033].\begin{bmatrix}1 & 3 & 0 & 2 & 1\\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 3 & 3\end{bmatrix}.

Step 3 — Swap R3,R4R_3,R_4; use new R3R_3 to clear column 4

After swap:

[13021001100003300000].\begin{bmatrix}1 & 3 & 0 & 2 & 1\\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 3 & 3\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}.

Divide R3R_3 by 3: (0,0,0,1,1)(0,0,0,1,1).

R2R2R3R_2\to R_2-R_3: (0,0,1,0,1)(0,0,1,0,-1).

R1R12R3R_1\to R_1-2R_3: (1,3,0,0,1)(1,3,0,0,-1).

[13001001010001100000].\begin{bmatrix}1 & 3 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 0\end{bmatrix}.

This is RREF.

Step 4 — Read rank

Three non-zero rows. Rank = 3.

Answer

  RREF as above; rank(A)=3.  \boxed{\;\text{RREF as above; rank}(A)=3.\;}
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