Maharashtra State BoardHSC Commerce 12th Board Exam
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Find the inverse of the following matrices by transformation method: [20-1510013] - Mathematics and Statistics

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Sum

Find the inverse of the following matrices by transformation method: `[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`

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Solution

Let A = `[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`

∴ |A| = `|(2, 0, -1),(5, 1, 0),(0, 1, 3)|`

= 2(3 – 0) –0(15 –0) –1(5 – 0)
= 6 – 0 – 5
= 1 ≠ 0
∴ A–1 exists.
Consider AA–1 = I

∴ `[(2, 0, -1),(5,1, 0),(0, 1, 3)] "A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Applying R1 ↔ R2, we get

`[(5, 1, 0),(2, 0, -1),(0, 1, 3)] "A"^-1 = [(0, 1, 0),(1, 0, 0),(0, 0, 1)]``

Applying R1 → R1 – 2R2, we get

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

Applying R2 → R2 – 2R1, we get

`[(1, 1, 2),(0, -2, -5),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(5, -2, 0),(0, 0, 1)]`

Applying R2 → R2 – 3R3, we get

`[(1, 1, 2),(0, 1, 4),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(5, -2, 0),(0, 0, 1)]`

Applying R1 → R1 – R2, R3 → R3 – R2,we get

`[(1, 0, -2),(0, 1, 4),(0, 0, -1)] "A"^-1 = [(-7, 3, -3),(5, -2, 3),(-5, 2, -2)]`

Applying R1 → (– 1) R3, we get

`[(1, 0, -2),(0, 1, 4),(0, 0, -1)] "A"^-1 = [(-7, 3, -3),(5, -2, 3),(-5, 2, -2)]`

Applying R1 → R1 + 2R3, R2 → R2 – 4R3, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`

∴ A–1 = `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`.

Concept: Inverse of Matrix
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