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

1 Answer
Mar 24, 2016

Answer:

#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#.