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If [(1,-1,1),(1,-1,1),(1,-1,1)], then A^...

If `[(1,-1,1),(1,-1,1),(1,-1,1)]`, then `A^5 - A^4 - A^3 + A^2`

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If A = [(2,2,1), (1,3,1), (1,2,2)] then A^-1+(A-5I) (AI)^2 = (i) 1/ 5 [[4,2, -1], [-1,3,1], [-1,2,4]] (ii) 1/5 [[4, -2, -1], [-1, 3, -1], [-1, -2,4]] (iii) 1/3 [[4,2, -1], [-1,3,1], [-1,2,4]] (iv) 1/3 [[4, -2, -1], [-1,3, -1], [-1, -2,4]]

if A=[{:(2,1,3),(1,-1,2),(4,1,5):}]and B=[{:(1,-1,2),(2,1,5),(4,1,3):}], then show that : (i) (A+B)'=A'+B' (ii) (A+4B)'=A'+4B'

Show that (i) [(5,-1),(6,7)][(2,1),(3,4)]!=[(2,1),(3,4)][(5,-1),(6,7)] (ii) [(1,2,3),(0,1,0),(1,1,1)][(-1,1,0),(0,-1,1),(2,3,4)]!=[(-1,1,0),(0,-1,1),(2,3,4)][(1,2,3),(0,1,0),(1,1,0)] .

show that (i) [{:(5,-1),(6,7):}][{:(2,3),(3,4):}]ne[{:(2,3),(3,4):}][{:(5,-1),(6,7):}] (ii) [{:(1,2,3),(0,1,0),(1,1,0):}][{:(-1,1,0),(0,-1,1),(2,3,4):}] ne[{:(-1,1,0),(0,-1,1),(2,3,4):}][{:(1,2,3),(0,1,0),(1,1,0):}]

The value of (1 - 1/2) (1 - 1/3) (1 - 1/4)(1 - 1/5) ………(1 - 1/n) is :

If A^(-1)=(1)/(3)[[1, 4, -2], [-2, -5, 4], [1, -2, 1]] and |A|=3 , then adjA=

Show that(i) [[5,-1],[ 6 ,7]][[2, 1],[ 3 ,4]]!=[[2, 1],[ 3, 4]][[5,-1],[ 6, 7]] (ii) [[1, 2, 3],[ 0, 1, 0],[ 1, 1, 0]][[-1, 1, 0],[ 0,-1, 1],[ 2, 3, 4]]!=[[-1, 1, 0],[ 0,-1, 1],[ 2, 3, 4]][[1 ,2 ,3],[ 0, 1, 0],[ 1, 1, 0]]

If A={:((3,1),(-4,5)):},B={:((-3,1),(4,-5)):}andC={:((2,1),(-1,3)):} , then

The value of (1-1/2)(1-1/3)(1-1/4)(1-1/5)…(1-1/n) is :