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You can ask questions about our review for exam 1 by posting them here!
MML Discourse archived in May, 2026
Questions on Review for Exam 1
User 009
The answers to number 6 seem to be:
a)
M=
\begin{bmatrix}
1 & 0 & 2 & 4 \\
0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}
b)
\begin{aligned}
x+2z = 4 \\
x = -2z+4 \\
\\
y+z = 2 \\
y = -z+2 \\
\\
z = z
\end{aligned}
❤️ 1
mark
Yes, I largely agree! Note that I did strike out your last equation z=z. I mean, it's hard to argue against that but the typical way that folks phrase that is to that "z is free".
I think the preferred way to write it is that the solution set is
\{(-2z+4, -z+2, z):z\in\mathbb R\}.
One final comment: On the exam itself, you show your work using notational steps like
\begin{bmatrix}
2 & 1 & 5 & 10\\
1 & -3 & -1 & -2\\
4 & -2 & 6 & 12
\end{bmatrix}
\xrightarrow{-2R_1+R_3}
\begin{bmatrix}
2 & 1 & 5 & 10\\
1 & -3 & -1 & -2\\
0 & -4 & -4 & -8
\end{bmatrix}.
I understand, of course, that typing that out in the forum is a skill we haven't really developed and am not expecting it here. I just want to make the expectations on the exam clear.
User 009
If z is free, can I just not write that and write the x and y equations as I've had them on the exam? I'm more referring to the format of the equation.
:thinking: 1
mark
Well, there's a whole infinite set of solutions, which why I often refer to the "solution set" and write it in set theoretic notation like
\{(-2z+4, -z+2, z):z\in\mathbb R\}.
If you'd like to incorporate your equations, you could write the solution as
The set of (x,y,z) such that
x=-2z+4 \text{ and } y = -z+2.
Again, the solution is not a set of equations; it's a set of points in 3D space defined by a set of equations.
User 009
I'll be using the second option then. That one is easier to remember.
👍 1
User 006
In regards to the definition question you will be asking us specifically on "Homogeneous System". Are you expecting us to remember the entire slide or the first sentence? I ask this clarifying question because the other key terms to remember had a solid one sentence definition that seemed suitable and the first sentence on this term seemed a bit vague for what you might expect on an exam.
❤️ 2
mark
Hmm... Yes, I agree that does look a bit vague. Let's just reorder things a bit:
An m\times n system A\vec{x}=\vec{b} is called homogeneous when \vec{b} is the zero vector \vec{b}=\vec{0}.
To be clear, that's a solid and complete definition using the matrix notation that we've developed since first meeting the concept. If you want to refer to a full system, you could write:
An m\times n system
\begin{aligned} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_m &= b_1 \\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_m &= b_2 \\ \vdots \hspace{5mm} + \hspace{5mm} \vdots \hspace{6mm} + \hspace{0.3mm} \ddots \hspace{0.5mm} + \hspace{6.7mm} \vdots \hspace{6mm} &= \hspace{2mm} \vdots \\ a_{m1} x_1 + a_{m2} x_2 + \cdots +a_{mn} x_m &= b_m \end{aligned}is called homogeneous if b_i = 0 for every i.
Either of the above formulations is fine.
👍 1
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