Question

How do you solve `5x+3y = -5` and `7x+5y = -11` using matrices?

Answer 1

`5x+3y=−5`
`7x+5y=−11`

Convert to matrix equation

`((5,3),(7,5)) ((x),(y))=((-5),(-11))`

Use definition of inverse to get
`A=((5,3),(7,5))`

`A^{-1}=1/{5*5-3*7}((5,-7),(-3,5))=1/4((5,-7),(-3,5))`

This becomes

`x=1/4*52=13`
`y=1/4*70=17.5`.

Explanation:

`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`