Isn’t the area of the parallelogram in the below case -13?
Consider the linear transformation T that maps the vectors (1,0) and (0,1) in the following manner:
T(0, 1) = (2, 5)
T(1, 0) = (3, 1)
The area of the parallelogram spanned by transforming the vectors (0,1) and (1,0) is :
Area cannot be negative.
So whats the answer? 13?
Well, determinants can be negative, but in this case it is not.
What is the T matrix that produces those outputs? What is its determinant?
The question was asked in the quiz. I don’t know the T matrix that produces those outputs. But when I typed in 13 as the answer, it gave me the wrong answer.
Well, they gave you enough information to figure out what the T matrix is, right? That’s the point. And once you know that, you can compute the determinant. No need to guess.
They told you that if you take T \cdot [[1], [0]] (where that notation means (1, 0) as a column vector) the answer is (3, 1), so that tells you what the first column of T is, right? And similarly with T \cdot [[0], [1]]. Think about how matrix multiplication works.
But vectors (2, 5) and (3, 1) are the transformed vectors of input vectors (0,1) and (1,0), isn’t it?
Yes, that’s the point. So that tells you what the linear transformation is. Do what I suggested: play out the linear transformation as a matrix multiply. Let’s say the matrix T is unknown, so call it [[a, b], [c, d]]. Now we know that T \cdot [[1], [0]] = [[3], [1]], so if you write out the result of the matrix multiply, that gives you two equations involving a and c, right?
BTW I did not find this question on the quiz in M4ML C1 W4. Is it embedded in one of the lectures? Please give me the name of the lecture and the time offset, if so.
Isn’t the area of a parallelogram defined as taking the determinant of the transformed vectors, in this case (2, 5) and (3, 1)?
Well, the area of the parallelogram is the determinant of the matrix representing the linear transformation. It is related to the values you show. Please read the other response I just gave you 2 minutes ago.
If I do as you suggested then -
[a b][0] = [2]
[c d][1] = [5]
[a b][1] = [3]
[c d][0] = [1]
provides -
[a] = [3] and [b] = [2]
[c] = [1] [d] = [5]
This means the T matrix is (3, 2) and (1, 5). However, if we take the determinant of this T matrix we still get 13.
This question is asked in the week 4 practice quiz of the Determinants-in-depth section.
I’m sorry for asking silly questions.
Ok, with your updated description, I found that quiz question. I gave the answer 13 and it was accepted by the grader.
There are two versions of this course. I wonder if one of you is enrolled in the older version.
I’m taking it as a normal paying customer, not as a mentor, so I don’t have the usual “Course Manager” tab. Is there a way for a normal student to get a “version number”?
Not that I’ve seen.
New students after the update were supposedly automatically enrolled in the new course.
Existing students stayed on the original course. The mechanism to ‘upgrade to the new course by resetting your deadlines’ apparently didn’t work.
Might be the only way to determine this is to compare the course content and look for differences.
The quiz question that Karan quoted looked exactly the same as what I’m seeing. It’s just that the answer is apparently different. I’m going to reserve judgement and suggest that Karan try it again and hope that this is just case of misinterpeting the symptoms. Note that you have to do “Try Again” and take the whole quiz again.