---
title: "Rank of a matrix by reduction [1,0,−5,6;3,−2,1,2;…]"
subject: "Mathematics III"
chapter: 1
key: "semester_3/mathematics_iii:rank-of-a-matrix-by-reduction-seven"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/rank-of-a-matrix-by-reduction-seven
site: BCT Engineering
---

# Rank of a matrix by reduction [1,0,−5,6;3,−2,1,2;…]

Mathematics III · Chapter 1: Determinants and Matrices

- Asked in: 2076 Chaitra, 5 marks
- Video: [Rank of the matrix (echelon form) problem part1. 5](https://www.youtube.com/watch?v=WxAXVYvM1o8) — Rajesh Lekkalu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/rank-of-a-matrix-by-reduction-seven
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1

## Question

Reduce the matrix [1 0 −5 6; 3 −2 1 2; 5 −2 −9 14; 4 −2 −4 8] into normal form and hence find its rank.

## Short answer

Reduce matrix by row operations: R₂−3R₁, R₃−5R₁, R₄−4R₁, then R₃−R₂, R₄−R₂. Further column operations yield the normal form with I₂. The rank is 2.

## Answer

```text
A = [1  0 −5  6]
    [3 −2  1  2]
    [5 −2 −9 14]
    [4 −2 −4  8]
```

**Step 1.** R₂ → R₂ − 3R₁, R₃ → R₃ − 5R₁, R₄ → R₄ − 4R₁:

```text
[1  0  −5    6]
[0 −2  16  −16]
[0 −2  16  −16]
[0 −2  16  −16]
```

**Step 2.** R₃ → R₃ − R₂, R₄ → R₄ − R₂:

```text
[1  0  −5    6]
[0 −2  16  −16]
[0  0   0    0]
[0  0   0    0]
```

**Step 3 (normal form).** Divide R₂ by −2: R₂ = (0, 1, −8, 8). Now column operations C₃ → C₃ + 5C₁ + 8C₂ and C₄ → C₄ − 6C₁ − 8C₂ wipe out the entries in columns 3 and 4:

```text
[1 0 0 0]
[0 1 0 0]   = [I₂ 0]
[0 0 0 0]     [0  0]
[0 0 0 0]
```

The normal form contains the identity matrix of order 2.

**Answer: rank(A) = 2.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which operations clear the first column?
   A. R2 -> R2 + 3R1, R3 -> R3 + 5R1, R4 -> R4 + 4R1
   B. R2 -> R2 - 3R1, R3 -> R3 - 5R1, R4 -> R4 - 4R1
   C. R2 -> R2 - R1 for all
   D. C2 -> C2 - 3C1
   Answer: B. Subtracting multiples of R1 removes 3, 5, 4. (Key point: First elimination)

2. After step 1, rows 2, 3 and 4 are all (0, -2, 16, -16). What does step 2 show?
   A. The rank is 4
   B. Two of them can be reduced to zero rows
   C. The matrix is singular only if row 1 is zero
   D. The rank is 1
   Answer: B. R3 - R2 and R4 - R2 give zero rows, so only two independent rows remain. (Key point: Dependent rows)

3. What is the normal form reached?
   A. I3 with a zero column
   B. [I2 0; 0 0]
   C. [I1 0; 0 0]
   D. The zero matrix
   Answer: B. Column operations wipe the remaining entries, leaving the identity of order 2. (Key point: Normal form)

4. The rank of A is:
   A. 1
   B. 4
   C. 3
   D. 2
   Answer: D. Rank equals the order of the identity in the normal form, which is 2. (Key point: Rank)

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