How do you solve #x+3y=5# and #x+4y=6# using matrices?

1 Answer
Mar 1, 2016

Answer:

From # vecR = A * vecx# multiplying both sides by #A^-1#
#A^-1*vecR = A^-1A*vecx = I * vecx #
where #A =((a_11,a_12), (a_21,a_22 )) =((1,3), (1,4 ))#
#vecx = ((x), (y)), vecR = ((5), (6))#
#A^-1 = ((4,-3 ), (-1, 1 )) #

Answere #=>((x), (y)) = ((2), (1)) #

Explanation:

Write,
#x+3y=5#
#x+4y=6#
#((1,3), (1,4 )) ((x), (y)) = ((5), (6))#
Let #A =((a_11,a_12), (a_21,a_22 )) =((1,3), (1,4 )), vecx = ((x), (y)), vecR = ((5), (6))#

"we need to find "
#A^-1# such that #I = A^-1 A " where " I = ((1,0 ), (0,1 ))# and we can write in vector matrix form:

# vecR = A * vecx# multiplying both sides by #A^-1#
#A^-1*vecR = A^-1A*vecx = I * vecx #
in our case:
#A^-1((5), (6)) = ((1,0 ), (0,1 )) ((x), (y))#
and the solution to #x, y#
#x = 5*a'_(11) + 6*a'_(12) #
#y = 5*a'_(21) + 6*a'_(22) #

but how do you find:
#A^-1 = ((a'_11, a'_12), (a'_21, a'_22))#?
#A^-1 = 1/det(A) * ((a_22, -a_12 ), (-a_21, a_11 ))# where
#det(A) = (a_11a_22 - a_12a_21) = 4-3= 1 #
Thus #A^-1 = ((4,-3 ), (-1, 1 )) #
#((x), (y)) =((4,-3 ), (-1, 1 )) ((5), (6)) =((2), (1)) #