4. Solve the following linear system using Gaussian elimination.
Show work.
3x + y – z = -5
-4x + y = 6
6x – 2y + 3z = 2
how about looking at
http://www.mathwords.com/g/gaussian_elimination.htm
x1+2x2+3x3=4
2x1-3x2+4x3-21=0
4x1+3x2-2x3+25=0
3 1 -1-5
-4 1 0 6
6 -2 3 2
=R3=r3+(-2r1)
=0 -4 1 12=R3
Sure, let's solve the linear system using Gaussian elimination. Here's the step-by-step process:
1. Write the augmented matrix:
[ 3 1 -1 | -5 ]
[-4 1 0 | 6 ]
[ 6 -2 3 | 2 ]
2. Perform row operations to eliminate the first and second variables below the first row:
[A1 - (A2 * (-4/3)), B1 - (B2 * (-4/3)), C1 - (C2 * (-4/3)), | D1 - (D2 * (-4/3))]
[ -4 1 0 | 6 ]
[ 6 -2 3 | 2 ]
[ 3 1 -1 | -5 ]
[ 0 5 -4 | -14]
[ 6 -2 3 | 2 ]
3. Perform row operations to eliminate the third variable below the second row:
[A1 - (A3 * 2), B1 - (B3 * 2), C1 - (C3 * 2), | D1 - (D3 * 2)]
[ 0 5 -4 | -14]
[ 6 -2 3 | 2 ]
[ 3 1 -1 | -5 ]
[ 0 5 -4 | -14]
[ 0 10 -9 | 12 ]
4. Perform row operations to eliminate the third variable below the first row:
[A1 - (A2 * (3/5)), B1 - (B2 * (3/5)), C1 - (C2 * (3/5)), | D1 - (D2 * (3/5))]
[ 3 1 -1 | -5 ]
[ 0 5 -4 | -14]
[ 0 10 -9 | 12 ]
[ 3 1 -1 | -5 ]
[ 0 5 -4 | -14]
[ 0 10 -9 | 12 ]
5. Perform row operations to eliminate the third variable below the second row:
[A2 - (A3 * 2), B2 - (B3 * 2), C2 - (C3 * 2), | D2 - (D3 * 2)]
[ 3 1 -1 | -5 ]
[ 0 5 -4 | -14]
[ 0 0 1 | -18 ]
6. Solve for the variables:
From the last row, we get z = -18.
Substituting the value of z into the second row equation:
5y - 4(-18) = -14
5y + 72 = -14
5y = -86
y = -86 / 5
Substituting the values of y and z into the first row equation:
3x + (-86/5) - (-18) = -5
3x - 86/5 + 18 = -5
3x - 86/5 = -5 + (90/5)
3x - 86/5 = 85/5
3x = (85/5) + (86/5)
3x = (85 + 86) / 5
3x = 171 / 5
x = 171 / (3 * 5)
x = 57 / 5
Therefore, the solution to the linear system is:
x = 57/5, y = -86/5, and z = -18.
To solve the linear system using Gaussian elimination, perform the following steps:
1. Write the augmented matrix for the system.
The augmented matrix for the given linear system is:
[ 3 1 -1 | -5 ]
[ -4 1 0 | 6 ]
[ 6 -2 3 | 2 ]
2. Perform row operations to simplify the matrix.
First, swap rows 1 and 2.
[ -4 1 0 | 6 ]
[ 3 1 -1 | -5 ]
[ 6 -2 3 | 2 ]
Next, multiply row 1 by -3/4 and add it to row 2.
[ -4 1 0 | 6 ]
[ 0 7 -1 | -8.5 ]
[ 6 -2 3 | 2 ]
Finally, multiply row 1 by -6/4 and add it to row 3.
[ -4 1 0 | 6 ]
[ 0 7 -1 | -8.5 ]
[ 0 -5 3 | -7 ]
3. Continue performing row operations to obtain row-echelon form.
To simplify further, multiply row 2 by 4/7 and add it to row 1.
[ -4 0 1/7 | 1.57 ]
[ 0 7 -1 | -8.5 ]
[ 0 -5 3 | -7 ]
Next, multiply row 2 by 5/7 and add it to row 3.
[ -4 0 1/7 | 1.57 ]
[ 0 7 -1 | -8.5 ]
[ 0 0 16/7 | -5.5 ]
Finally, divide row 3 by 16/7 to obtain a leading 1.
[ -4 0 1/7 | 1.57 ]
[ 0 7 -1 | -8.5 ]
[ 0 0 1 | -4.375 ]
4. Solve for the variables.
From the row-echelon form, we have the following equations:
-4x + (1/7)z = 1.57 (Equation 1)
7y - z = -8.5 (Equation 2)
z = -4.375 (Equation 3)
From Equation 3, z = -4.375.
Substituting z = -4.375 into Equation 1, we have:
-4x + (1/7)(-4.375) = 1.57
-4x - 0.625 = 1.57
-4x = 2.195
x = -0.5488
Finally, substituting z = -4.375 into Equation 2, we have:
7y - (-4.375) = -8.5
7y + 4.375 = -8.5
7y = -12.875
y = -1.8393
Therefore, the solution to the given linear system is:
x ≈ -0.5488
y ≈ -1.8393
z ≈ -4.375
You can also find a step-by-step solution on the website you provided: http://www.mathwords.com/g/gaussian_elimination.htm