If A = [[1,0,0],[-1,0,1],[0,1,0]], find the matrix A^3 calculated by the use of Cayley-Hamilton theorem or otherwise.

GATE 1993 · Engineering Mathematics · Matrix · medium

Answer: A^3 = [[1,0,0],[-2,0,1],[-1,1,0]]

1. Compute A^2 directly: A^2: row1=[1,0,0]*A = [1,0,0]; row2=[-1,0,1]*A = [-1+0, 0+1, 0+0] = [-1,1,0]; row3=[0,1,0]*A = [-1,0,1]. So A^2 = [[1,0,0],[-1,1,0],[-1,0,1]].
2. Find characteristic polynomial using Cayley-Hamilton: tr(A) = 1+0+0 = 1. det(A): expanding along row 1 = 1*(0*0-1*1) - 0 + 0 = -1. Sum of 2x2 principal minors: M11=det([[0,1],[1,0]])=-1, M22=det([[1,0],[0,0]])=0, M33=det([[1,0],[-1,0]])=0. Sum = -1. So p(lambda) = lambda^3 - lambda^2 - lambda + 1.
3. Compute A^3 = A^2 + A - I: A^2 + A = [[1+1,0+0,0+0],[-1-1,1+0,0+1],[-1+0,0+1,1+0]] = [[2,0,0],[-2,1,1],[-1,1,1]]. Then A^3 = [[2,0,0],[-2,1,1],[-1,1,1]] - [[1,0,0],[0,1,0],[0,0,1]] = [[1,0,0],[-2,0,1],[-1,1,0]].