How to do questions b and c?

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2 Answers
Feb 3, 2018

#"see explanation"#

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#

Feb 3, 2018

See below.

Explanation:

#bb(A)=[(2,1),(0,-1)]color(white)(88)##bb(B)=[(1,0),(3,1)]#

Find the product #bb(AB)#:

#bb(AB)=[(2,1),(0,-1)]*[(1,0),(3,1)]=[({2*1+1*3 }, {2*0+1*1}),({0*1-1*3},{0*0-1*1})]#

#=[(5,1),(-3,-1)]#

We can find the inverse of a #bb(2xx2)# matrix by switching the elements on the leading diagonal ( top left to bottom right ) and changing the sign of the elements on the non leading diagonal
( bottom left to top right ), and dividing by the determinant.

The determinant of a #bb(2xx2)# matrix:

#[(a,b),(c,d)]=(a*b)-(c*d)#

Determinant #bb(AB)#

#color(blue)(AB=[(5,1),(-3,-1)])#

#(5*-1)-(-3*1)=-2#

Inverse #bb(AB)#

Switch elements on leading diagonal:

#[(-1,1),(-3,5)]#

Change signs on non leading diagonal:

#[(-1,-1),(3,5)]#

Divide by the determinant:

#[(-1/-2,-1/-2),(3/-2,5/-2)]=[(1/2,1/2),(-3/2,-5/2)]#

#color(blue)(AB^-1=[(1/2,1/2),(-3/2,-5/2)])#

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.

#bb(A^-1)=[(1/2,1/2),(0,-1)]#

#bb(B^-1)=[(1,0),(-3,1)]#

#bb(A^-1B^-1)=[(-1,1/2),(3,-1)]#

#bb(B^-1A^-1)=[(1/2,1/2),(-3/2,-5/2)]#

This shows that:

#bb(B^-1A^-1)=bb((AB)^-1)#

This is a fundamental property, and is usually just expressed as:

For invertible matrices A and B

#bb((AB)^-1)=bb(B^-1A^-1)#