How do you solve #a+3b+2c=3# , #2a-b-3c= -8#, #5a+2b+c=9# using matrices?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Jan 19, 2018

Answer:

#a=2#, #b=-3# and #c=5#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,3,2,|,3),(2,-1,-3,|,-8),(5,2,1,|,9))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR2-2R1#; #R3larrR3-5R1#

#A=((1,3,2,|,3),(0,-7,-7,|,-14),(0,-13,-9,|,-6))#

#R2larr(R2)/(-7)#

#A=((1,3,2,|,3),(0,1,1,|,2),(0,-13,-9,|,-6))#

#R3larrR3+13R2#

#A=((1,3,2,|,3),(0,1,1,|,2),(0,0,4,|,20))#

#R3larr(R3)/4#

#A=((1,3,2,|,3),(0,1,1,|,2),(0,0,1,|,5))#

#R1larrR1-2R3#; #R2larrR2-R3#

#A=((1,3,0,|,-7),(0,1,0,|,-3),(0,0,1,|,5))#

#R1larrR1-3R2#

#A=((1,0,0,|,2),(0,1,0,|,-3),(0,0,1,|,5))#

Thus #a=2#, #b=-3# and #c=5#

Was this helpful? Let the contributor know!
1500