How do you solve #x+3y=11# and #x+4y=14# using matrices?

2 Answers
Jul 12, 2016

Answer:

x=2; y=3

Explanation:

you must solve by applying the basic propositions:

#x=D_x/D; y=D_y/D#

where D is the determinant of the matrix

#|quad1quad3|#
#|quad1 quad 4|#

Its determinant is obtained by:

#D=1*4-3*1=1#

#D_x and D_y# are respectively the determinants of the matrices

#|quad11quad3|#
#|quad14quad4|#

#|quad1quad11|#
#|quad1quad14|#

#D_x=11*4-3*14=44-42=2#

#D_y=1*14-11*1=3#

So

#x=D_x/D=2/1=2#

#y=D_y/D=3/1=3#

Jul 17, 2016

Answer:

#x=2" ; "y=3#

Two methods given

Explanation:

#color(blue)("Using Gauss Jordon elimination")#

#" "color(white)(.)xcolor(white)(.)ycolor(white)(.)"answer"#
#" "[[1,3,"|",11],[1,4,"|",14]]#
#" "Row2-Row1#
#" "darr#

#" "[[1,3,"|",11],[0,1,"|",3]]#
#" "Row1-3(Row2)#
#" "darr#

#" "[[1,0,"|",2],[0,1,"|",3]]" "larr [[x],[y]]#
#color(white)(.)#

#=>[[1,0],[0,1]] color(white)(.)[[x],[y]] = [[2],[3]]#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Using Linear Algebra")#

#A=[[1,3],[1,4]]#
#color(white)(.)#

#X=[[x],[y]]#

#B=[[11],[14]]#

#=>AX=B#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Consider the Algebra method of determining "x)#
Given:#" "3x=4#

Divide both side by 3 #-> 1/3xx3x=1/3xx4#

We multiply the 3 by 3 inverse to turn the coefficient of #x# into 1

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Using the Algebra method shown above")#

Multiply both sides by #A" inverse. written as "color(blue)(A^(-))#

#color(brown)(=>color(blue)(A^(-))AX" "=" "color(blue)(A^(-))B)#

#=>X=A^(-)B# ...................................Equation(1)

Checked in Maple:
Tony B
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "A^(-))#

#color(brown)("Method 1")#

Write the original matrix as:

#" "[[1,3,"|",1,0],[1,4,"|",0,1]]#

And conduct a Gauss Jordon elimination on the LHS

#" "Row2-Row1#
#" "darr#

#" "[[1,3,"|",1,0],[0,1,"|",-1,1]]#
#" "Row1-3(Row2)#
#" "darr#

#" "[[1,0,"|",4,-3],[0,1,"|",-1,1]] => A^(-)=[[4,-3],[-1,1]]#
,.....................................................................................
#color(brown)("Method 2 - Shortcut approach")#

Multiply a modified version of #[[1,3],[1,4]]# by its inverted determinant.

For #[[a,b],[c,d]]# the determinant is #ab-cd#
and the modified matrix is #[[d,-b],[-c,a]]#

#D=(1xx4)-(1xx3)=1#

Thus #A^(-)= 1xx[[4,-3],[-1,1]]#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Finish the calculation for Equation(1)")#
#=>X=A^(-)B# ...................................Equation(1)

#[[4,-3],[-1,1]] [[11],[14]] = [[2],[3]]#