Consider the matrices?

Question

Consider the matrices,3A=(3 -6), and, 1/2B=(1.5 3)
>>>>>>>>>>>>>>>>>>>> (9 0) >>>>>>>>> (0.5 -2)

(a) Find the matrices A and B, then compute the matrix, 0.5A – 3B.


(b) Find the matrices, A^–1, B^–1, (AB)^ –1, if possible.



(c)Compare det[(A – B)^3 ], with det[(B – A)^3]. Is this relation specific for only these matrices, or holds in general for all square matrices? Justify your answer.

Answer 1

(a)
To find matrix A, just divide each element by 3.
To find matrix B, just multiply each element by 2.

Then you should get
0.5A - 3B = [-8.5,-19 | -1.5,12]


(b)
A^-1 = [0,1/3 | -1/2,1/6]
B^-1 = [2/9,1/3 | 5/90,-1/6]


(c)
512 = 512
The condition holds in general.

Answer 2

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Answer 3

(a) A = 1/3 (3A) = :
1st row : 1 ; - 2
2nd row : 3 ; 0

B =
1st row : 3 ; 6
2nd row : 1 ; - 4

1/2A - 3B =
1st row : 0.5 - 9 ; - 1 - 18
2nd row : 1.5 - 3 ; 0 - 12

so :
1st row : - 8.5 ; - 19
2nd row : - 1.5 ; - 12

(b) det(A) = 1*0 - (-2)*3 = 6 then A^-1 exists

if A =
(a ; b)
(c ; d)

then A^-1 =
( d/det(A) ; -b/det(A) )
( -c/det(A) ; a/det(A) )

so A^-1 =
(0 ; 1/3)
(1/2 ; 1/6)

and B^-1 =
(2/9 ; 1/3)
(1/18 : -1/6)

(AB)^-1 = B^-1 * A^-1 =
(-1/6 ; 7/54)
(1/12 ; -1/108)

(c) det[(A - B)^3] = [det(A - B)]^3 = [- det(B - A)]^3
= - det[(B - A)^3]