How would you solve this matrices using the Gauss-jordan elimination method?
{[-5, 8, 10]}
{[5, 8, 4]}
{[10, 8, 5]}
{[5, 4, 8]}
-5 +5 10 +5
+8 +8 +8 +4
10 +4 +5 +8
divide row 1 by -5
+1 -1 -2 -1
+8 +8 +8 +4 subtract 8 -8 -16 -8
10 +4 +5 +8 subtract 10 =10 -20 -10
gives
+1 -1 -2 -1
+0 16 24 12
+0 14 25 18
divide row 2 by 16
+1 -1 -2 -1
+0 +1 1.5 .75 add to row 1-> new row 1
+0 14 25 18
gives
+1 0 -.5 -.25
+0 +1 1.5 .75 times 14 =0 14 21 21/2
+0 14 25 18
subtract row 2 times 14 from row 3
+1 0 -.5 -.25
+0 +1 1.5 .75
+0 +0 -4 -7.5 over 8 add to row 1
gives
+1 0 +0 +5.5/8
+0 +1 1.5 .75
+0 +0 -4 -7.5 divide by -4
then
+1 +0 +0 5.5/8
+0 +1 1.5 .75
+0 +0 +1 7.5/4 times 1.5, subtract from 2
then
+1 +0 +0 5.5/8
+0 +1 +0 -16.5/8
+0 +0 +1 7.5/4
so
x = 55/80 = .6875
y = -165/80 = -2.0625
z = 150/80 = 1.875
I did not bother to check all that, check my answers back in.
To solve a system of linear equations using the Gauss-Jordan elimination method, we need to perform elementary row operations on the augmented matrix until it is in row echelon form or reduced row echelon form.
The given system of linear equations can be represented as the augmented matrix:
[-5 8 10 | 0]
[ 5 8 4 | 0]
[10 8 5 | 0]
[ 5 4 8 | 0]
The goal is to transform this matrix into a form where the leading coefficients of each row are 1s, and all other entries in the corresponding columns are zeroes.
Here's a step-by-step procedure for solving the system of equations using Gauss-Jordan elimination:
Step 1: Swap rows if necessary to make the top-left entry non-zero.
Looking at the first column, we see that the top-left entry is already non-zero.
Step 2: Multiply the first row by a non-zero scalar, so that the leading coefficient becomes 1.
In this case, we can multiply the first row by -1/5:
[ 1 -8 -10 | 0]
[ 5 8 4 | 0]
[10 8 5 | 0]
[ 5 4 8 | 0]
Step 3: Perform row operations to eliminate all non-zero entries below the leading coefficient of the first row.
Using row operations, we can make the entries below the leading coefficient of the first row zero:
[ 1 -8 -10 | 0]
[ 0 12 14 | 0]
[ 0 12 15 | 0]
[ 0 12 18 | 0]
Subtracting the first row from each of the other rows multiplied by the appropriate scalar, we get:
[ 1 -8 -10 | 0]
[ 0 12 14 | 0]
[ 0 0 1 | 0]
[ 0 4 8 | 0]
Step 4: Proceed to the next column and repeat steps 1-3 for that column.
In the second column, the first non-zero entry is already 1. So, we move to the third column.
In the third column, the first non-zero entry is -10. We can divide the second row by 12/10 = 6/5 and subtract 14/10 = 7/5 times the first row from it. Similarly, we can divide the third row by 10/6 = 5/3 and subtract 15/6 = 5/2 times the first row from it. Finally, we divide the fourth row by 10/4 = 5/2 and subtract 8/4 = 2 times the first row from it.
After performing the necessary row operations, we have:
[ 1 -8 0 | 0]
[ 0 1 0 | 0]
[ 0 0 1 | 0]
[ 0 0 0 | 0]
Step 5: Convert the matrix to reduced row echelon form (RREF).
Since all leading coefficients are already 1, we can stop here. The last row implies 0x + 0y + 0z = 0, which is always true.
The RREF form of the augmented matrix corresponds to the system of equations:
x - 8y = 0
y = 0
z = 0
Therefore, the solution to the system of equations is x = 8y, y = 0, and z = 0.
In other words, any value of x that is 8 times the value of y will satisfy the given system of equations.