Question

How to do questions b and c?

Answer 1

`"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`

Answer 2

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)`