How do you solve the system #x+2y-z=6#, #-3x-2y+5z=-12#, and #x-2z=3#?
1 Answer
Please see the explantion
Explanation:
Write
Add a row for the equation
Add a row for the equation
Perform Elementary Row Operations until an identity matrix is obtained on the left.
We want the coefficient in position
We want the other two coefficients in column 1 to be 0, therefore, we perform the following 2 row operations:
Because Row 3 is a scalar multiple of Row 2, one should declare that there is no unique solution.
Let's write the first two rows as linear equations:
#2y+z=3
#y=3/2-z/2
Equations [1] and [2] will allow you to choose a value of x and then determine the corresponding value of y and z.