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Evaluate |(x, x+1),(x-1,x)|...

Evaluate `|(x, x+1),(x-1,x)|`

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`Let A = |(x,x+1),(x-1,x)|`
to find the determinant of A
`absA= x(x) - (x+1)(x-1)`
= `x^2 -[ x^2 -1]`
= `x^2 - x^2 +1`
= 1
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