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Question
If A = `[(1, 2, -1),(3, -2, 5)]`, apply R1 ↔ R2 and then C1 → C1 + 2C3 on A
Solution
A = `[(1, 2, -1),(3, -2, 5)]`
Applying R1 ↔ R2, we get
`[(3, -2, 5),(1, 2, -1)]`
Applying C1 → C1 + 2C3, we get
`[(13, -2, 5),(-1, 2, -1)]`
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