How do you solve the system #3a+2b=27, 6a-7b+c=5, -2a+10b+5c=-29# using matrices?

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Aug 18, 2017

Answer:

The solution is #((a),(b),(c))##=##((1323/199),(702/199),(-2029/199))#

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

#((3,2,0,:,27),(6,-7,1,:,5),(-2,10,5,:,-29))#

#R3larr3R3+R2#, #=>#, #((3,2,0,:,27),(6,-7,1,:,5),(0,23,16,:,-82))#

#R2larrR2-2R1#, #=>#, #((3,2,0,:,27),(0,-11,1,:,-49),(0,23,16,:,-82))#

#R2larr(R2)/(-11)#, #=>#, #((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,23,16,:,-82))#

#R3larr(R3)/(23)#, #=>#, #((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,1,16/23,:,-82/23))#

#R3larrR3-R2#, #=>#, #((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,0,199/253,:,-2029/253))#

#R3larr(R3)*(253/199)#, #=>#, #((3,2,0,:,27),(0,1,-1/11,:,49/11),(0,0,1,:,-2029/199))#

#R2larrR2+(1/11)R3#, #=>#, #((3,2,0,:,27),(0,1,0,:,702/99),(0,0,1,:,-2029/199))#

#RlarrR1-(2)R2#, #=>#, #((3,0,0,:,3969/199),(0,1,0,:,702/99),(0,0,1,:,-2029/199))#

#Rlarr(R1)/(3)#, #=>#, #((1,0,0,:,1323/199),(0,1,0,:,702/199),(0,0,1,:,-2029/199))#

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Aug 18, 2017

Answer:

#a=1323/199, b=702/199, c= -2029/199#

Explanation:

3a + 2b = 27
6a - 7b + c = 5
-2a + 10b +5c = -29

First, convert the left side of the equations into a matrix called 'A'. Inside the matrix, put in the coeffecients of each of the variables in the equations above.

A= #[(3,2,0),(6,-7, 1),(-2, 10, 5)]#

Now, do the same for right side of the equation and convert the numbers into a matrix called 'b'.

b= #[(27),(5),(-29)]#

Use the formula #Ax=b#, where #x# is the solution for the system of equations.

Rearrange for #x# #-># #x = A^-1 * b#

#A^-1# refers to the inverse of the #A# matrix. It can easily be found using a graphical calculator.

If you need to do it by hand, here are some videos from Khan Academy on how to do it. I did mine using Symbolab.

#x= [(3,2,0),(6,-7, 1),(-2, 10, 5)]^-1 * [(27),(5),(-29)]#

#x = [(1323/199),(702/199),(-2029/199)]#

Therefore,

#a=1323/199, b=702/199, c= -2029/199#

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