How do you solve using gaussian elimination or gauss-jordan elimination, #x+3y+z=7#, #x+y+4z=18#, #-x-y+z=7#?

Question

Benjamin Achtenberg · Accepted Answer

#x=-4#
#y=2#
#z=5# Initial Augmented Matrix: #[ ( 1.00, 3.00, 1.00, 7.00), ( 1.00, 1.00, 4.00, 18.00), ( -1.00, -1.00, 1.00, 7.00) ]# Pivot Action #n# pivot row = n; pivot column = n; pivot entry augmented matrix entry at (n,n) Pivot 1 Pivot Row 1 reduced by dividing all entries by 1.00 so pivot entry = 1 #[ ( 1.00, 3.00, 1.00, 7.00), ( 1.00, 1.00, 4.00, 18.00), ( -1.00, -1.00, 1.00, 7.00) ]# #[ ( 1.00, 3.00, 1.00, 7.00), ( 0.00, -2.00, 3.00, 11.00), ( 0.00, 2.00, 2.00, 14.00) ]# Pivot 2 Pivot Row 2 reduced by dividing all entries by -2.00 so pivot entry = 1 #[ ( 1.00, 3.00, 1.00, 7.00), ( 0.00, 1.00, -1.50, -5.50), ( 0.00, 2.00, 2.00, 14.00) ]# #[ ( 1.00, 0.00, 5.50, 23.50), ( 0.00, 1.00, -1.50, -5.50), ( 0.00, 0.00, 5.00, 25.00) ]# Pivot 3 Pivot Row 3 reduced by dividing all entries by 5.00 so pivot entry = 1 #[ ( 1.00, 0.00, 5.50, 23.50), ( 0.00, 1.00, -1.50, -5.50), ( 0.00, 0.00, 1.00, 5.00) ]# #[ ( 1.00, 0.00, 0.00, -4.00), ( 0.00, 1.00, 0.00, 2.00), ( 0.00, 0.00, 1.00, 5.00) ]#

Elias Ackerman · Answer

An alternate way to word the explanation follows a similar logic in solving for the row echelon form:

(reduced row echelon form would ask for #0#'s above all leading #1#'s as well.)

The augmented matrix is:

#[(1,3,1,|,7),(1,1,4,|,18),(-1,-1,1,|,7)]#

Now, operating on it, where the second indicated row number is the one that the operation is applied to, we get:

#stackrel(R_2 + R_3" ")(->)[(1,3,1,|,7),(1,1,4,|,18),(0,0,5,|,25)]#

#stackrel(-R_1 + R_2" ")(->)[(1,3,1,|,7),(0,-2,3,|,11),(0,0,5,|,25)]#

At this point we know enough to back-substitute via the Gaussian Elimination method.

#5z = 25 -> color(blue)(z = 5)#

#-2y + 3z = 11 -> -2y + 15 = 11 -> color(blue)(y = 2)#

#x + 3y + z = 7 -> x + 6 + 5 = 7 -> color(blue)(x = -4)#

Therefore, this method also gives you #color(blue)("("-4,2,5")")#.

Lincoln Anderson · Answer

undefined To solve the system of linear equations using Gaussian elimination or Gauss-Jordan elimination, follow these steps:

1. Write the system of equations in augmented matrix form:

$$ \begin{bmatrix} 1 & 3 & 1 & | & 7 \ 1 & 1 & 4 & | & 18 \ -1 & -1 & 1 & | & 7 \end{bmatrix} $$

2. Perform row operations to obtain a row-echelon form (or reduced row-echelon form) of the matrix. The goal is to get zeros below the main diagonal.

$$ \begin{bmatrix} 1 & 3 & 1 & | & 7 \ 0 & -2 & 3 & | & 11 \ 0 & 0 & 1 & | & 6 \end{bmatrix} $$

3. Use back substitution to find the values of the variables starting from the bottom row:

From the third row: $ z = 6 $

Substitute $ z = 6 $ into the second row to find $ y $:

$$ -2y + 3(6) = 11 $$
$$ -2y + 18 = 11 $$
$$ -2y = 11 - 18 $$
$$ -2y = -7 $$
$$ y = \frac{-7}{-2} = \frac{7}{2} $$

Substitute $ y = \frac{7}{2} $ and $ z = 6 $ into the first row to find $ x $:

$$ x + 3\left(\frac{7}{2}\right) + 1(6) = 7 $$
$$ x + \frac{21}{2} + 6 = 7 $$
$$ x + \frac{21}{2} = 1 $$
$$ x = 1 - \frac{21}{2} $$
$$ x = \frac{2}{2} - \frac{21}{2} $$
$$ x = \frac{-19}{2} $$

Therefore, the solution to the system of equations is $ x = \frac{-19}{2} $, $ y = \frac{7}{2} $, and $ z = 6 $.