#5x+3y=−5#
#7x+5y=−11#
#((5,3),(7,5)) ((x),(y))=((-5),(-11))#
Our linear system is now a matrix equation of the form.
#A v = c# with
#A=((5,3),(7,5))#
#v=((x),(y))# and
#c=((-5),(-11))#.
We want to find v, we need to rearrange this into something of the form #v=# something.
Just like solving any other equation we need to move #A# from the left hand side to the right. If a was a real number we'd do this my dividing both side by #A#, such that #1/A A v = c/A# this works because #1/A A=1#. #A# and #1/A# are inverses, they make 1, but what is 1 for a matrix and how can we find an inverse?
For matrices, 1 is the identify matrix, a matrix with 1's along the diagonal and zeros everywhere else, for a 2 by 2 matrix it is given by:
#I=((1,0),(0,1))#.
It is useful because I times any vector is just that vector, #Iv=v#, like #1 xx a=a#. A inverse of a matrix, #A^{-1}# is a matrix that when multiplied by A produces the identity matrix #A^{-1} A=I# or #A A^{-1} =I# . So solve #A v =c# we multiply both
side by the inverse of A.
#A^{-1} A v =A^{-1} c# since #A^{-1} A=I#
#I v =A^t c#
# v =A^t c#
At this point you just have to memorize something, the inverse of a 2 by 2 matrix.
If
#A=((a,b),(c,d))#
#A^-1=1/{ad-cb}((d,-c),(-b,a))#
Here
#A=((5,3),(7,5))#
#A^{-1}=1/{5*5-3*7}((5,-7),(-3,5))=1/4((5,-7),(-3,5))#
#((5,3),(7,5)) ((x),(y))=((-5),(-11))#
Multiple both side by the inverse we have
#1/4((5,-7),(-3,5))((5,3),(7,5)) ((x),(y))=1/4((5,-7),(-3,5))((-5),(-11))#
Using the definition of the inverse and some matrix multiplication, we are left with
#((1,0),(0,1)) ((x),(y))=1/4 (( 5*(-5) -7*(-11) ),( -3*(5)+ 5*(-11)))#
# ((x),(y))=1/4 ((-25+ 77),(-15-55))#
# ((x),(y))=1/4 ((52),(-70))#
# x=1/4*52=13#
# y=1/4*70=17.5#
