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1. |
`|[x+lambda, 2x, 2x], [2x, x+lambda, 2x], [2x, 2x, x+lambda]| =(5x+ lambda)(lambda-x)^(2)` |
Answer» Apply `R_(1) to R_(1) + R_(2) +R_(3) " and take" (5x + lambda) "common from "R_(1)` | |