given matrix A, find `alpha` if A^(n+1)=`alpha` A^n

A = (2 -1 -1)

(-1 2 -1)

(-1 -1 2)

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Since `alpha` cannot be a matrix, but a scalar, you need to find `A^2, A^3` , to identify the value of `alpha` , such that:

`A^2 = ((2,-1,-1),(-1,2,-1),(-1,-1,2))*((2,-1,-1),(-1,2,-1),(-1,-1,2))`

`A^2 =((6,-3,-3),(-3,6,-3),(-3,-3,6))`

Factoring out 3 yields:

`A^2 = 3*((2,-1,-1),(-1,2,-1),(-1,-1,2)) = 3*A`

Since `A^3 = A^2*A` , replacing `3*A` for `A^2` yields:

`A^3 = 3A*A => A^3 = 3A^2`

Comparing the result `A^3 = 3A^2` to the given condition `A^(n+1) = 3A^n` , yields that `alpha = 3` .

**Hence, evaluating the scalar `alpha` , under the given condition, yields `alpha = 3.` **

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