How do you solve # 6x+5y = 8# and #3x-y = 7# using matrices?

2 Answers

Answer:

#x=43/21# & #y=-6/7#

Explanation:

Given equations can be written as follows

#6x+5y-8=0\ ......(1)#

#3x-y-7=0\ ......(2)#

Using matrix method, the solution can be written as

#\frac{x}{(5)(-7)-(-8)(-1)}=\frac{y}{(-8)(3)-(6)(-7)}=\frac{1}{(6)(-1)-(5)(3)}#

#\frac{x}{-43}=\frac{y}{18}=-1/21#

#\implies \frac{x}{-43}=-1/21 #

#x=43/21#

#\implies \frac{y}{18}=-1/21 #

#y=-6/7#

Jul 22, 2018

Answer:

#color(blue)(x=43/21, y=-6/7)#

Explanation:

Put the coefficients of the equations into a matrix, and put the solutions into 1 column vector and the variables into another vector:

#bbA=[(6,5),(3,-1)]#

#bb(b)=[(8),(7)]#

#bbX=[(x),(y)]#

We now have the matrix equation:

#bb(AX)=bb(b)#

If #bbA# in invertible, then #bb(A^-1)# exists, so:

#bb(A^-1AX)=bb(A^-1)bb(b)#

#bb(IX)=bb(A^-1)bb(b)#

We need to find #bb(A^-1)#

First find the determinant of #bb(A)#

Determinant of #bbA=[(a,b),(c,d)]=ad-bc#

So determinant of:

#bbA=[(6,5),(3,-1)]=6(-1)-(5)(3)=-21#

Swap the elements on the leading diagonal and change the signs on the non-leading diagonal:

#bbA=[(-1,-5),(-3,6)]#

Next multiply by the reciprocal of the determinant:

#bb(A^-1)=[(1/21,5/21),(1/7,-6/21)]#

We now have:

#[(x),(y)]=[(1/21,5/21),(1/7,-6/21)][(8),(7)]=[(8/21+35/21),(8/7-2)]=[(43/21),(-6/7)]#

Hence:

#x=43/21#

#y=-6/7#