How do you find the inverse of #[(9,13), (27,36)]#?
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#A^(-1)=[(-4/3,13/27),(1,-1/3)]#
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)]#
The inverse is #=-1/27((36,-13),(-27,9))#
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))#
#A^-1=((-4/3,13/27),(1,-1/3))#
#"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))#
