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Let A=[(0,1),(0,0)], show that (aI+bA)^(...

Let `A=[(0,1),(0,0)]`, show that `(aI+bA)^(n)=a^(n)I+na^(n-1)bA`, where I is the identity matrix of order 2 and `n in N`.

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If I = [(1,0),(0,1)] and E = [(0,1),(0,0)] then show that (aI + bE)^(3) = a^(3)I+3a^(2)bE where I is identify matrix of order 2.

If I=[{:(1,0),(0,1):}] and E=[{:(0,1),(0,0):}] , show that (aI + bE)^(3) =a^(3)I + 3a^(2)bE . Where I is unit matrix of order 2.

Knowledge Check

  • If I_(n) is the identity matrix of order n then the rank of I_(n) is

    A
    1
    B
    `n+1`
    C
    no solution
    D
    `n-1`

  • If A=[(i,0,0),(0,i,0),(0,0,i)] then for ninN,A^(4n+1)=

    A
    `[(1,0,0),(0,1,0),(0,0,1)]`
    B
    `[(-1,0,0),(0,-1,0),(0,0,-1)]`
    C
    `[(i,0,0),(0,i,0),(0,0,i)]`
    D
    `[(-i,0,0),(0,-i,0),(0,0,-i)]`

  • The value of sum_(n= 1)^(13) (i^(n) + i^(n + 1)) , where i=sqrt(-1) is

    A
    `i`
    B
    `i-1`
    C
    `-i`
    D
    0

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    If A=[(i,0),(0,-i)],B=[(0,-1),(1,0)] and C=[(0,i),(i,0)] and I is the unit matrix of order 2, then show that (i) A^(2)=B^(2)=C^(2)=-I (ii) AB=-BA=-C .

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