How to do questions b and c?
2 Answers
Explanation:
(b)
AB=((2,1),(0,-1))((1,0),(3,1))
color(white)(AB)=((2+3,0+1),(0-3,0-1))=((5,1),(-3,-1))
"given a 2 by 2 matrix "A=((a,b),(c,d))
"then the "color(blue)"inverse matrix"
A^-1=1/(ad-bc)((d,-b),(-c,a))
• " if "ad-bc=0" then no inverse exists"
"for "AB
ad-bc=(5xx-1)-(1xx-3)=-2
"hence inverse matrix exists"
(AB)^-1=1/(-2)((-1,-1),(3,5))=((1/2,1/2),(-3/2,-5/2))
rArr(AB)^-1=1/2((1,1),(-3,-5))
(b)
"for matrix A"
ad-bc=-2
rArrA^-1=-1/2((-1,-1),(0,2))
"for marix B"
ad-bc=1
rArrB^-1=((1,0),(-3,1))
A^-1B^-1=-1/2((-1,-1),(0,2))((1,0),(-3,1))
color(white)(A^-1B^-1)=-1/2((-1+3,0-1),(0-6,0+2))
color(white)(A^-1B^-1)=-1/2((2,-1),(-6,2))=((-1,1/2),(3,-1))
B^-1A^-1=-1/2((1,0),(-3,1))((-1,-1),(0,2))
color(white)(B^-1A^-1)=-1/2((-1+0,-1+0),(3+0,3+2))
color(white)(B^-1A^-1)=-1/2((-1,-1),(3,5))=((1/2,1/2),(-3/2,-5/2))
A^-1B^-1!=B^-1A^-1rArr"non-commutative"
"and "(AB)^-1=B^-1A^-1
See below.
Explanation:
Find the product
We can find the inverse of a
( bottom left to top right ), and dividing by the determinant.
The determinant of a
Determinant
Inverse
Switch elements on leading diagonal:
Change signs on non leading diagonal:
Divide by the determinant:
We now need to find the the inverses of A and B.
I will just give you these. You can calculate them in exactly the same way as the previous one.
This shows that:
This is a fundamental property, and is usually just expressed as:
For invertible matrices A and B