Math Group
Question:
benjamin7
Student
Find the matrix that solves the equation. I is the identity matrix.
2 1 1 2
-1 3 . -1 2 F^T =28I
2 1 1 2
-1 3 is one matrix and -1 2 is another matrix and after them F^T and after thet equal to 28I (I is identity matrix).
Answers:
-
neela Teacher
Graduate SchooleNotes Editor
Posted by neela on Tuesday November 10, 2009 at 11:12 AMLet
A=[2 , 1
1 , 3] be a 2 by 2 matrix and
B=[1, 2
-1, 2] be another 2 by 2 matrix.
Then to find f such that A.B.F' = 28 I, where F' is the transpose of F.
A.B=
=[2-1,4+2
-1-3,-2+6] Or
A.B=
=[1 , 6
-4 , 4].
Let F' =
= [a , b
c , d]
Then A.B.F' is Product of
[1 , 6
-4 , 4]and
[a , b
c , d]
=[a+6c , b+6d
-4a+4c , -4b+4d]...........(1),which is equal to 28I
=[28 , 0
0 , 28].......................(2)
Equating the corresponding elements of matrices (1) and(2)
From the 1st column: a+6c=28......(3)and -4a+4c = 0...(4) gives c=a . Substituting in (3) we get: a+6a=28 or a=4 . Therefore c=4
From the 2nd column:
b+6d=0.....(5) and -4b+4d=28..........(3).
From (5) we get: b=-6d. plug this in (6): -4(-6d)+4d=28. So 28d=28 giving d = 1. Thtrefore b = -6d = -6.
Therefore , a=4, b=-6, c=4 and d=1
Therefore
F' = [a , b
c , d].
F= [a , c
b , d]
=[4 , 4
-6 , 1]
