If A=[(0 ,1 ,0),( 0, 0, 1),(p, q ,r)] an...

If `A=[(0 ,1 ,0),( 0, 0, 1),(p, q ,r)]` and `I` is the identity matrix of order 3, show that `A^3=p I+q A+r A^2` .

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Given,
`A=[(0 ,1 ,0),( 0, 0, 1),(p, q ,r)]`
Now,
`A^2=A*A`
`=>A^2=[(0 ,1 ,0),( 0, 0, 1),(p, q ,r)][(0 ,1 ,0),( 0, 0, 1),(p, q ,r)]`
`=[(0,0,1),(p,q,r),(rp,p+rq,q+r^2)]`
`A^3=A^2*A`
`=>A^3=[(0,0,1),(p,q,r),(rp,p+rq,q+r^2)][(0 ,1 ,0),( 0, 0, 1),(p, q ,r)]`
`=[(p,q,r),(rp,p+rq,q+r^2),(pq+r^2p,rp+q^2+r^2q,p+2rq+r^3)].....(1)`
`p I+q A+r A^2`
`=p[(1,0,0),(0,1,0),(0,0,1)]+q[(0 ,1 ,0),( 0, 0, 1),(p, q ,r)]+r[(0,0,1),(p,q,r),(rp,p+rq,q+r^2)]`
`=[(p,0,0),(0,p,0),(0,0,p)]+[(0 ,q ,0),( 0, 0, q),(pq, q^2 ,qr)]+[(0,0,r),(rp,rq,r^2),(r^2p,rp+r^2q,rq+r^3)]`
`=[(p,q,r),(rp,p+rq,q+r^2),(pq+r^2p,q^2+r^2p+rp,p+2qr+r^3)]....(2)`
From `(1)` and `(2)` we get,`A^3=p I+q A+r A^2`

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