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Question
Solve: `|(x,2,-1),(2,5,x),(-1,2,x)|` = 0
Solution
Expanding along R1 we get
`x|(5,x),(2,x)| - 2|(2,x),(-1,x)| - 1|(2,5),(-1,2)|` = 0
⇒ x(5x - 2x) - 2(2x + x) - 1(4 + 5) = 0
⇒ x(3x) - 2(3x) - 1(9) = 0
⇒ 3x2 - 6x - 9 = 0
⇒ x2 - 2x - 3 = 0 ....(Divided by 3)
⇒ (x - 3)(x + 1) = 0
⇒ x = 3 or -1
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