 In this video, we present the solution to question number 10 from the practice midterm exam for math 2270. We're given a matrix A, which is a three by three matrix, given as one, two, zero, zero, zero, one, and zero, zero, zero. And we're given a vector B in our three, three, one, zero. And we're asked to determine the general solution of the non-homogeneous system, AX equals B. And we're supposed to express it as a particular solution and the solution of the general solution of the homogeneous system in vector form. So to solve the system, we're gonna begin with A augment B, which for these coefficients, we're gonna get one, two, zero, slash three, zero, zero, one, one, and zero, zero, zero, zero. Which we can see that this matrix is already in REF. And so if it helps, we can think of the corresponding system of linear equations, right? This would say something like X1 plus 2X2 equals three and X3 equals one and then zero equals zero, which really doesn't help us at all. Why as we'll say the sky is blue in that situation, it's more helpful. So we know with our general solution, X right here, it can be broke up as X1, X2, X3. So what do we know? Well, X3 is a dependent variable. It has to be one, okay? We also know that X1 is a dependent variable. If you rewrite this first equation, we end up with X1 equals three minus 2X2. So we're gonna make that dependence statement right there. So X1 is three minus 2X2. And then, well, X2 doesn't have any restriction on it whatsoever. It's a free variable, so we get this. For which we can then decompose this very quickly. We see that a particular solution will be three, zero and one. So if X equals three, Y equals zero and Z are, yeah, if X1 equals three, X2 equals zero and X3 equals one, you can double check with these equations here, right? Three plus zero is equal to three and then one equals one, very trivial solution. And so this is gonna be our particular solution. Next, we're then gonna take the linear combination of X2 times what's left behind. We get a negative two, a one and a zero, like so, and so this is actually gonna be our general solution in play right here. Nothing more complicated than that. So X2 times negative two plus one, excuse me, negative two, one and zero there, that gives us the general solution of the homogeneous system. You'll notice that if you were to take this matrix and times it by negative two and negative two, one and zero, you would get doing the multiplication here. You're gonna get negative two plus two. You're gonna get a zero and a zero. This is the zero vector. Thus, this vector right here really does represent the null space, it's in the null space and that gives us a basis for the null space in fact. So we do have the general solution expressed as a particular solution and the general solution of the homogeneous system.