emmanck123
New member
- Joined
- Dec 7, 2020
- Messages
- 3
First of all, on monday I posted a similar post, but part of my exercise wasn't completed so I'm posting the completed version now. My problem with the exercise is that it generates 2 null rows. I asked my teacher and she finally responded, now my classes are in spanish so I had to translate what she told me,
"You can simplify both rows, simplifying them with another row reduction, that is to say, both null rows are proposed as a new matriz( the wording here is nuevo escalonamiento, which translated directly would be new row reduction, this is the only part im not sure I translated properly) to be solved by row operations to reduce it, and the last row is simplified. The simplified row is then used to solve for IMT(image)."
From what I understood, I took the values obtained from my null rows, R3= a/3-b/3+c and R4=-a/3-4/3b+d. And made both into a new matrix. The problem is that I get a unique solution, which would mean my range would be 4 (Its a linear transformation from R3 to R4) and if I add both dimensions, NT+IMT=1+4=5, which is wrong, since I should get 3. Unless I misunderstood, I believe thats how I should solve it. I'm also thinking that the values I obtain from reducing R4 are the ones I should use for finding the image, in which case my range would be 2, which added with NT, should give me 3.
"You can simplify both rows, simplifying them with another row reduction, that is to say, both null rows are proposed as a new matriz( the wording here is nuevo escalonamiento, which translated directly would be new row reduction, this is the only part im not sure I translated properly) to be solved by row operations to reduce it, and the last row is simplified. The simplified row is then used to solve for IMT(image)."
From what I understood, I took the values obtained from my null rows, R3= a/3-b/3+c and R4=-a/3-4/3b+d. And made both into a new matrix. The problem is that I get a unique solution, which would mean my range would be 4 (Its a linear transformation from R3 to R4) and if I add both dimensions, NT+IMT=1+4=5, which is wrong, since I should get 3. Unless I misunderstood, I believe thats how I should solve it. I'm also thinking that the values I obtain from reducing R4 are the ones I should use for finding the image, in which case my range would be 2, which added with NT, should give me 3.
