How do you solve #2x - 4y = 7# and #-3x + y = 12# using matrices?

1 Answer

Answer:

#x= -5.5#
#y= - 4.5#

Explanation:

From the given equations

#2x-4y=7" "#first equation
#-3x+y=12" "#second equation

Equations take the form:

#a_11x+b_12y=c_13#
#a_21x+b_22y=c_23#

So that we have
#a_11=2# and #a_21=-3#
#b_12=-4# and #b_22=1#
#c_13=7# and #c_23=12#

The formulas

#Delta=[(a_11, b_12),(a_21,b_22)]=a_11*b_22-a_21*b_12#

#x=([(c_13, b_12),(c_23,b_22)])/([(a_11, b_12),(a_21,b_22)])=(c_13*b_22-c_23*b_12)/(a_11*b_22-a_21*b_12)#

#y=([(a_11,c_13),( a_21,c_23)])/([(a_11, b_12),(a_21,b_22)])=(a_11*c_23-a_21*c_13)/(a_11*b_22-a_21*b_12)#

Let us solve for x and y

#Delta=[(a_11, b_12),(a_21,b_22)]=a_11*b_22-a_21*b_12#

#Delta=[(2, -4),(-3,1)]=2*1-(-3)*(-4)=2-12=-10#

#x=([(c_13, b_12),(c_23,b_22)])/([(a_11, b_12),(a_21,b_22)])=(c_13*b_22-c_23*b_12)/(a_11*b_22-a_21*b_12)#

#x=([(7, -4),(12,1)])/([(2, -4),(-3,1)])=(7*1-12*-4)/(2*1-(-3)*(-4))=(7+48)/(-10)=55/-10#

#x=-5.5#

#y=([(a_11,c_13),( a_21,c_23)])/([(a_11, b_12),(a_21,b_22)])=(a_11*c_23-a_21*c_13)/(a_11*b_22-a_21*b_12)#

#y=([(2,7),( -3,12)])/([(2, -4),(-3,1)])=(2*12-(-3)*(7))/(2*1-(-3)*(-4))=(24+21)/(-10)=45/(-10)#

#y=-4.5#

God bless...I hope the explanation is useful.