solve the system of equation for all variables using a matrix
2x+y+z=6
-3x-4y+2z=4
x=y-z=-2
To solve the system of equations using a matrix, we can represent the system in matrix form and then use matrix operations to find the solution.
Let's start by writing the system of equations in matrix form:
[2 1 1 | 6]
[-3 -4 2 | 4]
[1 -1 -1 | -2]
In this matrix, each row represents an equation with the coefficients of the variables. The last column represents the constants on the right side of each equation.
To solve the system, we need to perform row operations on the matrix until it is in reduced row-echelon form using techniques such as Gaussian elimination or Gauss-Jordan elimination.
Performing the row operations:
1. Start with the first row and divide it by 2 to make the leading coefficient (the number in the first column) equal to 1:
[1 1/2 1/2 | 3]
2. Multiply the first row by 3 and add it to the second row to eliminate the x variable in the second row:
[1 1/2 1/2 | 3]
[0 -7/2 5/2 | -5]
3. Multiply the first row by -1 and add it to the third row to eliminate the x variable in the third row:
[1 1/2 1/2 | 3]
[0 -7/2 5/2 | -5]
[0 -3/2 -3/2 | -5]
4. Multiply the second row by -2/7 to make the leading coefficient of the second row equal to 1:
[1 1/2 1/2 | 3]
[0 1 -5/7 | 10/7]
[0 -3/2 -3/2 | -5]
5. Multiply the second row by -1/2 and add it to the first row to eliminate the y variable in the first row:
[1 0 4/7 | 4/7]
[0 1 -5/7 | 10/7]
[0 -3/2 -3/2 | -5]
6. Multiply the second row by 3/2 and add it to the third row to eliminate the y variable in the third row:
[1 0 4/7 | 4/7]
[0 1 -5/7 | 10/7]
[0 0 0 | 0]
Now the matrix is in reduced row-echelon form.
We can read the solution from the matrix:
x = 4/7
y = 10/7
z can be any value as indicated by the zero row
Therefore, the system of equations is consistent and has infinitely many solutions. The solutions can be represented as an infinite number of ordered triples (x, y, z). In this case, it is (4/7, 10/7, z).