How do you solve the system of equations: \begin{array}{ l }{ 3x + 2y + 4z = 11} \\ { 2x - y + 3z = 4} \\ { 5x - 3y + 5z = - 1} \end{array}?

Enter your answer as an ordered triple.

1 Answer
Douglas K.
Jan 2, 2017

Verify that the determinant is not zero; if so, then write an augmented matrix and perform elementary row operations, until you obtain an identity matrix on the left.

Explanation:

Verify that the Determinant is not zero:

#|(3,2,4),(2,-1,3),(5,-3,5)| = 18#

The system has a unique solution.

Write the Augmented Matrix

#[ (3,2,4,|,11), (2,-1,3,|,4), (5,-3,5,|,-1) ]#

Perform Elementary Row Operations until you obtain an Identity Matrix on the left:

#R_1 - R_2 to R_1#

#[ (1,3,1,|,7), (2,-1,3,|,4), (5,-3,5,|,-1) ]#

#R_2 - 2R_1 to R_2#

#[ (1,3,1,|,7), (0,-7,1,|,-10), (5,-3,5,|,-1) ]#

#R3 - 5R_1 to R_3#

#[ (1,3,1,|,7), (0,-7,1,|,-10), (0,-18,0,|,-36) ]#

#R_2 harr R_3#

#[ (1,3,1,|,7), (0,-18,0,|,-36), (0,-7,1,|,-10) ]#

#(-1/18)R_2 to R_2#

#[ (1,3,1,|,7), (0,1,0,|,2), (0,-7,1,|,-10) ]#

#R_3 + 7R_2 to R_3#

#[ (1,3,1,|,7), (0,1,0,|,2), (0,0,1,|,4) ]#

#R_1 - R_3 to R_1#

#[ (1,3,0,|,3), (0,1,0,|,2), (0,0,1,|,4) ]#

#R_1 - 3R_2 to R_1#

#[ (1,0,0,|,-3), (0,1,0,|,2), (0,0,1,|,4) ]#

We have an identity matrix on the left, therefore, we can read the solution set on the right:

#x = -3, y = 2, and z =4#

Check:

#3(-3)+2(2)+4(4) =11#
#2(-3)−(2)+3(4)=4#
#5(-3)−3(2)+5(4)=−1#

#11 =11#
#4=4#
#−1=−1#

This checks

The ordered triple is #(-3,2,4)#

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