# Find the inverse of the following matrices by transformation method: [20-1510013] - Mathematics and Statistics

Sum

Find the inverse of the following matrices by transformation method:

[(2, 0, −1),(5, 1, 0),(0, 1, 3)]

#### 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)]

By R1→3R1,

[(6, 0, –3),(5, 1, 0),(0, 1, 3)] "A"^-1 = [(3, 0, 0),(0, 1, 0),(0, 0, 1)]

By R1 → R1 − R2

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

By R2 → R2 − 5R1

[(1, –1, –3),(0, 6, 15),(0, 1, 3)] "A"^-1 = [(3, –1, 0),(–15, 6, 0),(0, 0, 1)]

By R2↔R3

[(1, –1, –3),(0, 1, 3),(0, 6, 16)] "A"^-1 = [(3, –1, 0),(0, 0, 1),(–15, 6, 0)]

By R1→R1 + R2 and R3→R3 − 6R2

[(1, 0, 0),(0, 1, 3),(0, 0, –3)] "A"^-1 = [(3, –1, 1),(0, 0, 1),(–15, 6, –6)]

By R3 → (- 1)/3 "R"_3,

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

By R2 →R2 – 3R3

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

IA–1 = [(3, –1, 1),(–15, 6, –5),(5, –2, 2)]

A–1 = [(3, –1, 1),(–15, 6, –5),(5, –2, 2)]

Is there an error in this question or solution?
Chapter 2: Matrices - Exercise 2.5 [Page 72]

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