Question

How do you solve `4x-y=-3` and `5x-3y=-23` using matrices?

Answer 1

`x=2 and y=11`

Explanation:

There are 3 ways using matrices and linear algebra :

Method number 1 - Gauss-Jordan Elimination

Write the linear system in the form `Ax=b` where

`A=[(4,-1),(5,-3)]` is the coefficient matrix of the system,

`x=[(x),(y)]` is the column vector of variables, and

`b=[(-3),(-23)]` is the column vector of solutions.

Now augment the coefficient matrix with the column vector of solutions and perform elementary row operations until you obtain the `2xx2` identity matrix `I_2` on the left hand side and then what remains on the right hand side will be the solution set for `x`. I will attach a sketch at the end to show these steps.
Eventually we get :

`[(4,-1),(5,-3)]|[(-3),(-23)] -> [(1,0),(0,1)]|[(2),(11)]`

and hence `x=2 and y=11`.

Method number 2 - Using the inverse matrix method

Since `Ax=b` it implies that `x=A^(-1)b`.

We may find `A^(-1)` by using elementary row operations in augmenting the coefficient matrix A with the identity matrix `I_2`, alternatively, we may use the formula :
`A^(-1)=1/(det(A))*adj(A)`, where `det(A)=Delta`.

`=1/(-12-(-5))*[(-3,1),(-5,4)]`

`=[(3/7,-1/7),(5/7,-4/7)]`

Hence `x=A^(-1)b=[(3/7,-1/7),(5/7,-4/7)][(-3),(-23)]=[(2),(11)]`

and so `x=2 and y=11`.

Method number 3 - Kramer's Rule

Find the determinants of the matrices formed by replacing each column vector in the coefficient matrix with the column vector of solutions :

`Delta_x=|(-3,-1),(-23,-3)|=(-3)(-3)-(-1)(-23)=-14`.

`Delta_y=|(4,-3),(5,-23)|=-77`.

Then : `x=Delta_x/Delta=-14/-7=2`

and `y=Delta_y/Delta=-77/-7=11`.