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