use matrices to solve the system of equations , if possible.
2x+y+2z=4
2x+2y=5
2x-y+6z=2
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
2 1 2 4
2 2 0 5
2-1 6 2
1 0 2 3/2
0 1-2 1
0 0 0 0
(3/2, 1, 0)
To solve the system of equations using matrices, we can represent the system in the form of an augmented matrix. Let's start by writing down the equation system:
2x + y + 2z = 4 -----> 2x + y + 2z = 4
2x + 2y = 5 -------> 2x + 2y = 5
2x - y + 6z = 2 -----> 2x - y + 6z = 2
Now, let's write the augmented matrix by putting the coefficients of the variables (x, y, z) and the constants on the right-hand side in a matrix form:
[ 2 1 2 | 4 ]
[ 2 2 0 | 5 ]
[ 2 -1 6 | 2 ]
The above matrix representation can be denoted as [A | B], where A represents the matrix of the coefficients of the variables and B represents the column matrix of the constants on the right-hand side.
To solve the system using matrices, we will perform row operations on the matrix [A | B] to transform it into an upper triangular matrix.
The goal is to use row operations to transform the matrix into the following form:
[ a b c | d ]
[ 0 e f | g ]
[ 0 0 h | i ]
Once we reach this form, we can easily solve for the variables starting from the bottom row.
Now, let's perform the row operations step by step:
1. Row 2 = Row 2 - Row 1
2. Row 3 = Row 3 - Row 1
The new matrix will be:
[ 2 1 2 | 4 ]
[ 0 1 -2 | 1 ]
[ 0 -2 4 | -2 ]
3. Row 3 = Row 3 + 2 * Row 2
The updated matrix will be:
[ 2 1 2 | 4 ]
[ 0 1 -2 | 1 ]
[ 0 0 0 | 0 ]
Now, the matrix is in upper triangular form. We can easily solve for the variables starting from the bottom row:
From the bottom row, we can see that 0z = 0, which means z can take any value (it is a free variable).
From the second row, we have 1y - 2z = 1. We substitute the value of z with a parameter (let's say t) to get y:
y - 2t = 1 -----> y = 2t + 1
Finally, from the first row, we have 2x + y + 2z = 4. Substituting the values of y and z obtained above:
2x + (2t + 1) + 2t = 4
2x + 4t + 1 = 4
2x + 4t = 3
x + 2t = 3/2
x = 3/2 - 2t
So, the solution to the system of equations is:
x = 3/2 - 2t
y = 2t + 1
z = t
Note that since z is a free variable, the system has infinitely many solutions.