Question

How do you find the inverse of `[(9,13), (27,36)]`?

Answer 1

`A^(-1)=[(-4/3,13/27),(1,-1/3)]`

Explanation:

Let's say `A=[(9,13),(27,36)]`

The inverse is then : `A^(-1)=1/(det(A))adj(A)`

`1/(det(A))=1/(9*36-13*27)=-1/27`

`adj(A)=[(36,-13),(-27,9)]`

So `A^(-1)=-1/27[(36,-13),(-27,9)]=[(-36/27,13/27),(27/27,-9/27)]=`

`[(-4/3,13/27),(1,-1/3)]`

Answer 2

The inverse is `=-1/27((36,-13),(-27,9))`

Explanation:

The inverse of the matrix `A=((a,b),(c,d))` is

`A^-1=1/(detA)*((d,-b),(-c,a))`

Let `A=((9,13),(27,36))`

The determinant of `A` is

`detA=|((9,13),(27,36))|=36*9-27*13=-27`

As `detA!=0`, the matrix `A` is invertible

`A^-1=-1/27((36,-13),(-27,9))`

Verification

`A*A^-1=((9,13),(27,36))*(-1/27)((36,-13),(-27,9))=((1,0),(0,1))`

Answer 3

`A^-1=((-4/3,13/27),(1,-1/3))`

Explanation:

`"given " A=((a,b),(c,d))`

`"then the inverse "A^-1" is"`

`A^-1=1/(detA)((d,-b),(-c,a))`

`detA=ad-bc`

`rArrdetA=(9xx36)-(13xx27)=-27`

`rArrA^-1=-1/27((36,-13),(-27,9))`

`color(white)(rAeeA^-1)=((-4/3,13/27),(1,-1/3))`