If #A=[(-1,2), (3,1)]#, how do you find F(A) where #f(x)=x^2-2x+3#?

2 Answers
Mar 7, 2017

Answer:

#2 [(6 , -2),(-3 , 4)]#

See below.

Explanation:

Write it as:

#f(A)=[(-1,2), (3,1)]cdot[(-1,2), (3,1)] - 2[(-1,2), (3,1)] + 3 I#

Where #I# is the identity matrix: #[(1,0), (0,1)]#

And then process the algebra.

If you need an answer check, I get:

#f(A) = 2 [(6 , -2),(-3 , 4)]#

Mar 8, 2017

Answer:

#f(A) = A^2-2A+3I = ((12,-4),(-6,8))#

Explanation:

Given:

#A = ((-1, 2),(3, 1))#

Then:

#A^2 = ((-1, 2),(3, 1))((-1, 2),(3, 1)) = ((7, 0),(0, 7)) = 7I#

So:

#A^2-2A+3I = 7I-2A+3I#

#color(white)(A^2-2A+3I) = 10I-2A#

#color(white)(A^2-2A+3I) = ((10,0),(0,10))-((-2,4),(6,2))#

#color(white)(A^2-2A+3I) = ((12,-4),(-6,8))#

#color(white)()#
Footnote

Note that #A# is a "square root of #7#". That is, it is a root of the polynomial:

#x^2-7 = 0#

As a result, we find that the set of matrices of the form #pI+qA# where #p# and #q# are rational is a field under matrix addition and multiplication. This field is essentially the same as (i.e. isomorphic to) the set of numbers of the form #p+qsqrt(7)# for rational multipliers #p, q#.

Note that this matrix #A# is not the only possible way to represent #sqrt(7)# with a rational matrix. For example, you could use #((0, 7),(1, 0))#.