1.

If `A=diag [a,b,c]` then show that `A^n=diag[a^n,b^n,c^n]` for all `n in N.`

Answer» We shall prove the result by mathematical induction.
When n=1, we have
`A^(1)=" diag "[a^(1),b^(1),c^(1)]=" diag "[a, b, c]=A.`
So, the result is true for n=1.
Let it be true for n=m, so that
`A^(m)=" diag "[a^(m),b^(m),c^(m)]" "...(i)`
`:." "A^(m+1)=A.A^(m)`
`=" diag "[a,b,c]." diag "[a^(m),b^(m),c^(m)]`[using (i)]
`=[{:(a,0,0),(0,b,0),(0,0,c):}].[{:(a^(m),0,0),(0,b^(m),0),(0,0,c^(m)):}]`
`=[{:(a^(m+1),0,0),(0,b^(m+1),0),(0,0,c^(m+1)):}]=" diag "[a^(m+1),b^(m+1),c^(m+1)].`This shows that the result is true for `n=(m+1)`, whenever it is true for n=m,
Hence, by the principle of mathematical iduction, the result is true for all `n inN.`


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