system of equations:
solve the following system of operations using matrices (row operations)
{5x + y + z + w = -5}
{x - 2y + 2z -3w = -4}
2 equations
4 unknowns
no single solution.
I need more than that. What are the values of "x"
I worked out the values of "y" to be:
y = (15/11) + (9/11z) - (16/11w), but I could be wrong
oh, well, if that's the kind of solution you want, then it's ok.
Eliminating x, I get
5x+y+z+w = -5
5x-10y+10z-15w = -20
11y-9z+16w = 15
so your answer looks good to me.
so are you saying that "x" is 15?
No. I used your answer and showed that no matter what x,z,w are, the expression you got will work for y. We eliminated x, remember?
To solve this system of equations using matrices and row operations, we can set up an augmented matrix.
First, we can write the coefficients of the variables as the elements of the matrix. The augmented matrix represents the left side of the equations along with the constants on the right side. The row operations will then help us transform the augmented matrix into row echelon form or reduced row echelon form.
The augmented matrix for this system of equations is:
[5 1 1 1 | -5]
[1 -2 2 -3 | -4]
Now let's perform the row operations to transform the augmented matrix into row echelon form.
1. R1 <- R1/5:
Divide the first row by 5 to make the leading coefficient in the first row a 1.
[1/5 1/5 1/5 1/5 | -1]
[1 -2 2 -3 | -4]
2. R2 <- R2 - R1:
Subtract the first row from the second row to eliminate the x variable.
[1/5 1/5 1/5 1/5 | -1]
[4/5 -11/5 9/5 -14/5 | -3]
3. R2 <- R2 * 5:
Multiply the second row by 5 to make the leading coefficient in the second row a 1.
[1/5 1/5 1/5 1/5 | -1]
[4/5 -11/5 9/5 -14/5 | -3]
4. R1 <- R1 - R2/4:
Subtract one-fourth of the second row from the first row to eliminate the x variable.
[1/5 0 1/10 3/10 | 0]
[4/5 -11/5 9/5 -14/5 | -3]
5. R2 <- R2 + 11R1/4:
Add eleven-fourths of the first row to the second row to eliminate the y variable.
[1/5 0 1/10 3/10 | 0]
[0 -1 5/2 -1 | -3]
Now we have obtained the row echelon form. To find the values of the variables, we can back-substitute.
From the second equation, we have -y + (5/2)z - w = -3. Rearranging this equation, we get -y = -3 - (5/2)z + w, which can be written as y = 3 + (5/2)z - w.
From the first equation, using this value of y, we have (1/5)x + (1/10)z + (3/10)w = 0.
Rearranging this equation, we get x = - (1/10)z - (3/10)w.
Therefore, the solutions to the system of equations are:
x = - (1/10)z - (3/10)w
y = 3 + (5/2)z - w
z = z (a free variable)
w = w (a free variable)
Using this method, we can solve the given system of equations using matrices and row operations.