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fi A=[(1,1),(1,1)] then A^(3)=...

fi `A=[(1,1),(1,1)]` then `A^(3)=`

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If A=[(1,-1),(-1,1)], "then" " "A^(3)

If A=[(1,1,1),(1,1,1),(1,1,1)] then show that A^n=[(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1))] .

If A=[(1,2,1),(0,1,-1),(3,-1,1)] then A^(3)-3A^(2)-A+9I=

If A=[(1,1,1),(1,1,1),(1,1,1)] , prove that A^(n)=[(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1))],n in N .

If A=[(1,1,1),(1,1,1),(1,1,1)] , prove that A^(n)=[(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1))],n in N .

If A=[(1,1,1),(1,1,1),(1,1,1)] , prove that A^(n)=[(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1))],n inN

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=([1,2,1],[0,1,-1],[3,-1,1]) then A^(3)-3A^(2)-A-9=