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)