Find the inverse of matrix A = [101023121] by using elementary row transformations - Mathematics and Statistics

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Sum

Find the inverse of matrix A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` by using elementary row transformations 

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Solution

A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` 

∴ |A| = `|(1, 0, 1),(0, 2, 3),(1, 2, 1)|` 

= 1(2 – 6) – 0 + 1(0 – 2)

= 1(– 4) + 1(– 2)

= – 4 – 2

= – 6 ≠ 0

∴ A–1 exists.

Consider AA–1 = I

∴ `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` A–1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying R3 → R3 – R1, we get

`[(1, 0, 1),(0, 2, 3),(0, 2, 0)]` A–1 = `[(1, 0, 0),(0, 1, 0),(-1, 0, 1)]`

Applying R2 → `(1/2)` R2, we get

`[(1, 0, 1),(0, 1, 3/2),(0, 2, 0)]` A–1 = `[(1, 0, 0),(0, 1/2, 0),(-1, 0, 1)]`

Applying R3 → R3 – 2R2, we get

`[(1, 0, 1),(0, 1, 3/2),(0, 0, -3)]` A–1 = `[(1, 0, 0),(0, 1/2, 0),(-1, -1, 1)]`

Applying R3 → `(-1/3)` R3, we get

`[(1, 0, 1),(0, 1, 3/2),(0, 0, 1)]` A–1 = `[(1, 0, 0),(0, 1/2, 0),(1/3, 1/3, -1/3)]`

Applying R1 → R1 – R3, R2 → R2 – `(3/2)` R3, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` A–1 = `[(2/3, -1/3, 1/3),(-1/2, 0, 1/2),(1/3, 1/3, -1/3)]`

∴ A–1 = `[(2/3, -1/3, 1/3),(-1/2, 0, 1/2),(1/3, 1/3, -1/3)]`

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Chapter 1.2: Matrices - Q.5

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