---
title: "Gauss-Jordan inverse of matrix"
subject: "Numerical Method"
chapter: 3
key: "semester_4/numerical_method:gauss-jordan-inverse-3x3"
source: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-jordan-inverse-3x3
site: BCT Engineering
---

# Gauss-Jordan inverse of matrix

Numerical Method · Chapter 3: Solution of system of linear algebraic equations

- Asked in: 2073 Magh, 8 marks
- Video: [How to find the inverse of matrix || Gauss Joradn method || 3*3 matrix || Hindi || Mathematics](https://www.youtube.com/watch?v=81LjGqfpdJg) — ENGINEER'S ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/gauss-jordan-inverse-3x3
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3

## Question

Compute the inverse of the following matrix using the Gauss-Jordan Method.
[3 1 2; 1 2 3; 2 3 5]

## Short answer

Gauss-Jordan method on [A|I] yields inverse after row operations. A⁻¹ = [0.5000 0.5000 -0.5000; 0.5000 5.5000 -3.5000; -0.5000 -3.5000 2.5000].

## Answer

**Gauss-Jordan method for the inverse:** write the augmented matrix [A | I] and apply row operations until the left half becomes the identity matrix I. The right half is then A⁻¹.

#### Step 1: Check that the inverse exists
det A = 3(2·5 − 3·3) − 1(1·5 − 3·2) + 2(1·3 − 2·2) = 3(1) − 1(−1) + 2(−1) = 2 ≠ 0, so A⁻¹ exists.

#### Step 2: Row operations
Augmented matrix [A | I]:

```text
3.0000  1.0000  2.0000  |  1.0000  0.0000  0.0000
1.0000  2.0000  3.0000  |  0.0000  1.0000  0.0000
2.0000  3.0000  5.0000  |  0.0000  0.0000  1.0000
```

R1 ← R1 / 3.0000:

```text
1.0000  0.3333  0.6667  |  0.3333  0.0000  0.0000
1.0000  2.0000  3.0000  |  0.0000  1.0000  0.0000
2.0000  3.0000  5.0000  |  0.0000  0.0000  1.0000
```

R2 ← R2 − (1.0000) R1; R3 ← R3 − (2.0000) R1:

```text
 1.0000   0.3333   0.6667  |   0.3333   0.0000   0.0000
 0.0000   1.6667   2.3333  |  -0.3333   1.0000   0.0000
 0.0000   2.3333   3.6667  |  -0.6667   0.0000   1.0000
```

R2 ← R2 / 1.6667:

```text
 1.0000   0.3333   0.6667  |   0.3333   0.0000   0.0000
 0.0000   1.0000   1.4000  |  -0.2000   0.6000   0.0000
 0.0000   2.3333   3.6667  |  -0.6667   0.0000   1.0000
```

R1 ← R1 − (0.3333) R2; R3 ← R3 − (2.3333) R2:

```text
 1.0000   0.0000   0.2000  |   0.4000  -0.2000   0.0000
 0.0000   1.0000   1.4000  |  -0.2000   0.6000   0.0000
 0.0000   0.0000   0.4000  |  -0.2000  -1.4000   1.0000
```

R3 ← R3 / 0.4000:

```text
 1.0000   0.0000   0.2000  |   0.4000  -0.2000   0.0000
 0.0000   1.0000   1.4000  |  -0.2000   0.6000   0.0000
 0.0000   0.0000   1.0000  |  -0.5000  -3.5000   2.5000
```

R1 ← R1 − (0.2000) R3; R2 ← R2 − (1.4000) R3:

```text
 1.0000   0.0000   0.0000  |   0.5000   0.5000  -0.5000
 0.0000   1.0000   0.0000  |   0.5000   5.5000  -3.5000
 0.0000   0.0000   1.0000  |  -0.5000  -3.5000   2.5000
```

#### Step 3: Result
The left half is I, so

A⁻¹ =
```text
 0.5000   0.5000  -0.5000
 0.5000   5.5000  -3.5000
-0.5000  -3.5000   2.5000
```

In fractions: A⁻¹ = ½ × [1 1 −1; 1 11 −7; −1 −7 5].

Check: row 1 of A times column 1 of A⁻¹ = 3(0.5) + 1(0.5) + 2(−0.5) = 1, and row 1 of A times column 2 of A⁻¹ = 3(0.5) + 1(5.5) + 2(−3.5) = 0, as required for A·A⁻¹ = I.

**Answer: A⁻¹ = [0.5000 0.5000 -0.5000; 0.5000 5.5000 -3.5000; -0.5000 -3.5000 2.5000]**

## Check yourself

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

1. In the Gauss-Jordan method for the inverse, what is done to [A | I]?
   A. Row operations turn the left half into zeros
   B. Row operations turn the left half into I, and the right half becomes A^-1
   C. Columns are swapped until the right half is A
   D. A is multiplied by its determinant
   Answer: B. Reducing [A | I] to [I | A^-1] gives the inverse on the right. (Key point: Method)

2. What check shows A^-1 exists for A = [3 1 2; 1 2 3; 2 3 5]?
   A. det A = 2, which is not zero
   B. det A = 0
   C. A is symmetric
   D. All entries are positive
   Answer: A. An inverse exists only when det A ≠ 0; here det A = 3(1) - 1(-1) + 2(-1) = 2. (Key point: Existence of inverse)

3. What is the second row of A^-1?
   A. [0.5, 0.5, -0.5]
   B. [-0.5, -3.5, 2.5]
   C. [1, 11, -7]
   D. [0.5, 5.5, -3.5]
   Answer: D. The final right half is ½ × [1 1 -1; 1 11 -7; -1 -7 5], so its second row is [0.5, 5.5, -3.5]. (Key point: Result)

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