Use matrices to solve the system. (If the system has infinitely many solutions, express your answer in terms of c, where x = x(c), y = y(c), and z = c. If the system has no solution, enter NONE for each answer.)
5x+2y-z=2
x-2y+2z= -1
3y+z= -17
ive done tons of these but this one is giving me a bit of trouble.. please help!
Is your augmented matrix correct ?
5 2 -1 | 2
1 -2 2 | -1
0 3 1 | -17
I don't know which matrix method you are supposed to use.
e.g. Gauss-Jordan ? or .....
hard to line up matrices nicely on this forum.
yes but i still don't get the answer!!!
x=1, y=-4, z=-5
To solve this system of equations using matrices, we can use the augmented matrix method. We'll create an augmented matrix and perform row operations to put the matrix in row-echelon form or reduced row-echelon form. Let's begin:
Step 1: Create the augmented matrix by writing down the coefficients and the constants on the right-hand side of the equations:
[5 2 -1 | 2]
[1 -2 2 | -1]
[0 3 1 | -17]
Step 2: Perform row operations to simplify the matrix, keeping in mind the goal is to reach row-echelon form or reduced row-echelon form.
a) Replace Row2 with Row2 - 5 * Row1:
Row2: [0 -12 12 | -6]
b) Replace Row3 with Row3 - (3/5) * Row1:
Row3: [0 3 1 | -17] - (3/5) * [5 2 -1 | 2]
Row3: [0 3 1 | -17] - [3 6 -3 | 6]
Row3: [0 3 1 | -17] - [3 6 -3 | 6]
Resulting matrix after step 2:
[5 2 -1 | 2]
[0 -12 12 | -6]
[0 0 0 | -23]
Step 3: Analyze the resulting matrix:
The third row has all zeros on the left side but a non-zero constant on the right side, -23. This indicates that the system has no solution.
Therefore, the system of equations has no solution. We can express this by assigning "NONE" to each variable:
x = NONE
y = NONE
z = NONE