Andrew E. answered • 22d
Mechanical Engineering Ph.D. Student Helping Students Master Algebra 2
Hello Steve,
When you are presented with a system of linear equations in matrix form, the three equations can be written as such: 2X+2Y+0Z=0, PX+2Y+0Z=0, QX-1Y-1Z=0. There are a few solution methods you can use. When it comes to three equations, I typically like to use the Elimination method. This method lines up the three equations above one another, and attempts to manipulate 1 or more equations at a time so that when the equations are added together, the coefficients cancel each other out. However, I believe the Substitution method works better in this situation since in the first two equations, the coefficient in front of "Z" is 0.
The substitution method works by isolating 1 variable in an equation and solving it in terms of another. We will use this on the first equation. You can use any equation, but I am choosing equation 1 because the coefficient in front of the "Z" is 0.
First, I choose to solve for X in equation 1. This gets us X = -Y.
Here comes the substitution part. We take this new equation for X and substitute in equation 2 for the variable X. This works because these equations are assumably linearly dependent. This means we are assuming that they cross each other at 1 x,y coordinate on a curve. When you substitute, you get
P(-Y)+2Y+0Z=0. This turns into Y(-P+2)=0, solving for Y we get 0.
Because our goal is to solve for P, we can rearrange this so that we get -P+2=0, therefore, P=2.
Now that we know that Y=0, we can plug Y back into the equation determined for X, X = -Y. This makes X=0 as well.
Knowing X and Y, we plug these values into equation 3, solving for Z. This gets us, Q(0)-1(0)-1(Z)=0. This gets us Z=0.
This tells us that, given the current info, it is not possible to determine the value of Q. No equation in the system provides a constraint involving Q that would allow us to solve for it, so Q remains undetermined.
In summary: P=2, Q=Undetermined
Hope this helps!
-Andrew Evans
