Find the unique solutions for the systems of linear equations
w+2x+2y-z=5
2w-x-y+4z=5
-w+3x+4y-2z=5
3w+x+y-4z=4
To find the unique solutions for this system of linear equations, we can use the method of elimination or substitution. However, since it involves four variables, elimination might be more efficient. Here's how we can proceed:
First, let's rewrite the system of equations in matrix form:
| 1 2 2 -1 | | w | | 5 |
| 2 -1 -1 4 | * | x | = | 5 |
| -1 3 4 -2 | | y | | 5 |
| 3 1 1 -4 | | z | | 4 |
Now, we can perform row operations to get the system of equations into row-echelon form. We'll start by getting a leading 1 in the first column of the first row. To do this, we'll swap the first and third rows:
| -1 3 4 -2 | | y | | 5 |
| 2 -1 -1 4 | * | x | = | 5 |
| 1 2 2 -1 | | w | | 5 |
| 3 1 1 -4 | | z | | 4 |
Next, we'll add 2 times the first row to the second row, and 3 times the first row to the fourth row:
| -1 3 4 -2 | | y | | 5 |
| 0 5 7 2 | * | x | = | 15 |
| 1 2 2 -1 | | w | | 5 |
| 0 10 13 -8 | | z | | 19 |
Finally, we'll add the first row to the third row:
| -1 3 4 -2 | | y | | 5 |
| 0 5 7 2 | * | x | = | 15 |
| 0 5 6 -3 | | w | | 10 |
| 0 10 13 -8 | | z | | 19 |
Now, we can subtract one-fifth times the second row from the third row:
| -1 3 4 -2 | | y | | 5 |
| 0 5 7 2 | * | x | = | 15 |
| 0 0 -1 -1 | | w | | 5 |
| 0 10 13 -8 | | z | | 19 |
Next, we'll multiply the third row by -1:
| -1 3 4 -2 | | y | | 5 |
| 0 5 7 2 | * | x | = | 15 |
| 0 0 1 1 | | w | | -5 |
| 0 10 13 -8 | | z | | 19 |
Finally, we'll subtract two times the third row from the fourth row:
| -1 3 4 -2 | | y | | 5 |
| 0 5 7 2 | * | x | = | 15 |
| 0 0 1 1 | | w | | -5 |
| 0 10 11 -10 | | z | | 21 |
Now, the system is in row-echelon form. By performing back substitution, we can solve for the variables:
From the fourth row, we have 10z - 11w = 21. Rearranging, we get z = (21 + 11w) / 10.
From the third row, we have w + z = -5. Substituting the expression for z, we get w + (21 + 11w) / 10 = -5. Multiplying both sides by 10, we get 10w + 21 + 11w = -50. Simplifying, we have 21w = -71. Thus, w = -71 / 21.
From the second row, we have 5x + 7y + 2z = 15. Substituting the expressions for x and z, we get 5x + 7y + 2((21 + 11w) / 10) = 15. Multiplying both sides by 10, we get 50x + 70y + 42 + 22w = 150. Simplifying, we have 50x + 70y + 22w = 108.
From the first row, we have -x + 3y + 4z = 5. Substituting the expressions for x and z, we get -((21 + 11w) / 10) + 3y + 4((21 + 11w) / 10) = 5. Multiplying both sides by 10, we get -21 - 11w + 30y + 42 + 22w = 50. Simplifying, we have 30y + w = 29.
Therefore, the unique solution to the system of linear equations is:
w = -71/21
x = 1
y = 1
z = (21 + 11w) / 10.
To find the unique solutions for the system of linear equations, we can use the method of Gaussian elimination or matrix operations. Let's solve the system step by step using Gaussian elimination:
Step 1: Write the augmented matrix of the system:
[ 1 2 2 -1 | 5 ]
[ 2 -1 -1 4 | 5 ]
[-1 3 4 -2 | 5 ]
[ 3 1 1 -4 | 4 ]
Step 2: Perform row operations to transform the augmented matrix into row-echelon form (or reduced row-echelon form). We'll start with the first column:
R2 = R2 - 2R1
R3 = R3 + R1
R4 = R4 - 3R1
The new augmented matrix:
[ 1 2 2 -1 | 5 ]
[ 0 -5 -5 6 | -5 ]
[ 0 5 6 -3 | 10 ]
[ 0 -5 -5 1 | -11 ]
Step 3: Continue performing row operations to eliminate the coefficients below the diagonal:
R3 = R3 + R2
R4 = R4 + R2
The new augmented matrix:
[ 1 2 2 -1 | 5 ]
[ 0 -5 -5 6 | -5 ]
[ 0 0 1 3 | 5 ]
[ 0 0 0 7 | -16 ]
Step 4: Perform backward substitutions to find the values of the variables. Start with the last row:
z = -16/7
Substituting z back into the third row equation:
w + 2x + 2y - (-16/7) = 5
w + 2x + 2y + 16/7 = 5
w + 2x + 2y = 5 - 16/7
w + 2x + 2y = 35/7 - 16/7
w + 2x + 2y = 19/7
Simplifying:
w + 2x + 2y = 19/7
Now, let's substitute the value of z in the second row equation:
-5x - 5y + 6(-16/7) = -5
-5x - 5y - 96/7 = -35/7
-5x - 5y = -35/7 + 96/7
-5x - 5y = 61/7
Simplifying:
-5x - 5y = 61/7
And, substituting z in the first row equation:
w + 2x + 2y - (-16/7) = 5
w + 2x + 2y + 16/7 = 5
w + 2x + 2y = 5 - 16/7
w + 2x + 2y = 35/7 - 16/7
w + 2x + 2y = 19/7
Now we have the system of equations:
w + 2x + 2y = 19/7
-5x - 5y = 61/7
w + 2x + 2y = 19/7
We can simplify the second equation by dividing by -5:
x + y = -61/35
From these equations, we can see that the variables w, x, and y are not uniquely determined. There are infinitely many solutions for this system of linear equations.