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" (ii) "-1.0-1.5-2.0-2.5-....=-27...

" (ii) "-1.0-1.5-2.0-2.5-....=-27

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If A={:[(3,2),(0,1)]:}" then:(A^(-1))^(3)= ...a) (1)/(27)[(1,-26),(0,27)] b) (1)/(27)[(-1,26),(0,27)] c) (1)/(27)[(1,-26),(0,-27)] d) (1)/(27)[(-1,-26),(0,-27)]

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The inverse of the matrix [(1,1,1),(1,0,2),(3,1,1)] is a) (1)/(4)[(-2,0,2),(5,-1,2),(1,-1,-2)] b) (1)/(4)[(-2,0,2),(5,-2,-1),(1,2,-1)] c) (1)/(4)[(-2,0,2),(2,5,-1),(-2,-1,1)] d) (1)/(4)[(-2,0,2),(5,-1,1),(1,-2,-1)]

Simplify: [(2 10/27))^(-2/3) ÷ (11 1/9))^(-0.5)] +[(6.25)^(0.5) ÷ (-4)^(-1)]

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if A=[{:(3,-1,2),(0,5,-3),(1,-2,7):}]and B=[{:(1,0,0),(0,1,0),(0,0,1):}], find whether AB=BA or Not .