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