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Question
Evaluate the determinant.
`|(3,-1,-2),(0,0,-1),(3,-5,0)|`
Solution
|A| = `abs ((3,-1,-2),(0,0,-1),(3,-5,0))`
`= 3|(0,-1),(-5,0)| + 1|(0,-1),(3,0)| - 2|(0,0),(3,-5)|`
`= 3 [0 xx 0 - (- 1) xx (-5)] + 1[0 xx 0 - (- 1) xx 3] - 2[0 xx (- 5) - 0 xx (3)]`
= 3[0 - 5] + 1[0 + 3] - 2 [0 - 0]
= 3 × (- 5) + 1[0 + 3] - 2[0 - 0]
= - 15 + 3
= - 12
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