Row equivalence and solutions
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Let Axb S and A'xb' S' be two systems of m linear s in n-unknowns. Suppose the exted matrix A'|b' is row equivalent to A|b. Then LS'LS.
Solution:
Proof. We with the following claim: bf claim : If S' can be obtained from S by one elementary row operation then LS'subseteq LS. bf Proof of the first claim: Let x_...x_nin LS. we need to show x_...x_nin LS'. itemize item operation : c R_i rightarrow R_i c neq . All the s of S' coincide with the s of S except of i a_ix_+...+a_inx_n b_i longrightarrow c a_ix_+...+c a_inx_n c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the first operation. item operation : R_j+c R_i longrightarrow R_j. Again all the s of S and S' coincide except of j. a_jx_+...+a_jnx_n b_j longrightarrow a_jx_+...+a_jnx_n b_j+c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the second operation. item operation : R_i longleftrightarrow R_j. In this case we clearly have x_...x_nin LS Longrightarrow x_...x_nin LS' because the order of the s in the system does not have any effect on the set of solutions. itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then S can be obtained from S' by one elementary row operation. bf Proof of the second claim: Again there are three cases. itemize item operation : S rightarrow S' c R_i rightarrow R_i But then S' rightarrow S R_i rightarrow fraccR_i item operation : S rightarrow S' R_j+c R_i rightarrow R_j quad ineq j But then S' rightarrow S R_j-c R_i rightarrow R_j item operation : S rightarrow S' R_i leftrightarrow R_j But then S' rightarrow S R_j leftrightarrow R_i itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then LSLS'. bf Proof of the third claim: By bf claim LS subseteq LS'. By bf claim S is obtained from S' by one elementary row operation hence by bf claim again with the roles of S and S' reversed we have LS' subseteq LS. This concludes the proof of bf claim . We are now in position to prove the Theorem from the ning. By asption there is a finite sequence of elementary row operations SS_ rightarrow S_ rightarrow ... rightarrow S_kS'. By bf claim LSLS_LS_...LS_kLS'.
Let Axb S and A'xb' S' be two systems of m linear s in n-unknowns. Suppose the exted matrix A'|b' is row equivalent to A|b. Then LS'LS.
Solution:
Proof. We with the following claim: bf claim : If S' can be obtained from S by one elementary row operation then LS'subseteq LS. bf Proof of the first claim: Let x_...x_nin LS. we need to show x_...x_nin LS'. itemize item operation : c R_i rightarrow R_i c neq . All the s of S' coincide with the s of S except of i a_ix_+...+a_inx_n b_i longrightarrow c a_ix_+...+c a_inx_n c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the first operation. item operation : R_j+c R_i longrightarrow R_j. Again all the s of S and S' coincide except of j. a_jx_+...+a_jnx_n b_j longrightarrow a_jx_+...+a_jnx_n b_j+c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the second operation. item operation : R_i longleftrightarrow R_j. In this case we clearly have x_...x_nin LS Longrightarrow x_...x_nin LS' because the order of the s in the system does not have any effect on the set of solutions. itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then S can be obtained from S' by one elementary row operation. bf Proof of the second claim: Again there are three cases. itemize item operation : S rightarrow S' c R_i rightarrow R_i But then S' rightarrow S R_i rightarrow fraccR_i item operation : S rightarrow S' R_j+c R_i rightarrow R_j quad ineq j But then S' rightarrow S R_j-c R_i rightarrow R_j item operation : S rightarrow S' R_i leftrightarrow R_j But then S' rightarrow S R_j leftrightarrow R_i itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then LSLS'. bf Proof of the third claim: By bf claim LS subseteq LS'. By bf claim S is obtained from S' by one elementary row operation hence by bf claim again with the roles of S and S' reversed we have LS' subseteq LS. This concludes the proof of bf claim . We are now in position to prove the Theorem from the ning. By asption there is a finite sequence of elementary row operations SS_ rightarrow S_ rightarrow ... rightarrow S_kS'. By bf claim LSLS_LS_...LS_kLS'.
Meta Information
Exercise:
Let Axb S and A'xb' S' be two systems of m linear s in n-unknowns. Suppose the exted matrix A'|b' is row equivalent to A|b. Then LS'LS.
Solution:
Proof. We with the following claim: bf claim : If S' can be obtained from S by one elementary row operation then LS'subseteq LS. bf Proof of the first claim: Let x_...x_nin LS. we need to show x_...x_nin LS'. itemize item operation : c R_i rightarrow R_i c neq . All the s of S' coincide with the s of S except of i a_ix_+...+a_inx_n b_i longrightarrow c a_ix_+...+c a_inx_n c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the first operation. item operation : R_j+c R_i longrightarrow R_j. Again all the s of S and S' coincide except of j. a_jx_+...+a_jnx_n b_j longrightarrow a_jx_+...+a_jnx_n b_j+c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the second operation. item operation : R_i longleftrightarrow R_j. In this case we clearly have x_...x_nin LS Longrightarrow x_...x_nin LS' because the order of the s in the system does not have any effect on the set of solutions. itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then S can be obtained from S' by one elementary row operation. bf Proof of the second claim: Again there are three cases. itemize item operation : S rightarrow S' c R_i rightarrow R_i But then S' rightarrow S R_i rightarrow fraccR_i item operation : S rightarrow S' R_j+c R_i rightarrow R_j quad ineq j But then S' rightarrow S R_j-c R_i rightarrow R_j item operation : S rightarrow S' R_i leftrightarrow R_j But then S' rightarrow S R_j leftrightarrow R_i itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then LSLS'. bf Proof of the third claim: By bf claim LS subseteq LS'. By bf claim S is obtained from S' by one elementary row operation hence by bf claim again with the roles of S and S' reversed we have LS' subseteq LS. This concludes the proof of bf claim . We are now in position to prove the Theorem from the ning. By asption there is a finite sequence of elementary row operations SS_ rightarrow S_ rightarrow ... rightarrow S_kS'. By bf claim LSLS_LS_...LS_kLS'.
Let Axb S and A'xb' S' be two systems of m linear s in n-unknowns. Suppose the exted matrix A'|b' is row equivalent to A|b. Then LS'LS.
Solution:
Proof. We with the following claim: bf claim : If S' can be obtained from S by one elementary row operation then LS'subseteq LS. bf Proof of the first claim: Let x_...x_nin LS. we need to show x_...x_nin LS'. itemize item operation : c R_i rightarrow R_i c neq . All the s of S' coincide with the s of S except of i a_ix_+...+a_inx_n b_i longrightarrow c a_ix_+...+c a_inx_n c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the first operation. item operation : R_j+c R_i longrightarrow R_j. Again all the s of S and S' coincide except of j. a_jx_+...+a_jnx_n b_j longrightarrow a_jx_+...+a_jnx_n b_j+c b_i If x_...x_n satisfies the left part then it satisfies also the right one. This proves the claim for the second operation. item operation : R_i longleftrightarrow R_j. In this case we clearly have x_...x_nin LS Longrightarrow x_...x_nin LS' because the order of the s in the system does not have any effect on the set of solutions. itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then S can be obtained from S' by one elementary row operation. bf Proof of the second claim: Again there are three cases. itemize item operation : S rightarrow S' c R_i rightarrow R_i But then S' rightarrow S R_i rightarrow fraccR_i item operation : S rightarrow S' R_j+c R_i rightarrow R_j quad ineq j But then S' rightarrow S R_j-c R_i rightarrow R_j item operation : S rightarrow S' R_i leftrightarrow R_j But then S' rightarrow S R_j leftrightarrow R_i itemize This concludes the proof of bf claim . bf claim : If S' is obtained from S by one elementary row operation then LSLS'. bf Proof of the third claim: By bf claim LS subseteq LS'. By bf claim S is obtained from S' by one elementary row operation hence by bf claim again with the roles of S and S' reversed we have LS' subseteq LS. This concludes the proof of bf claim . We are now in position to prove the Theorem from the ning. By asption there is a finite sequence of elementary row operations SS_ rightarrow S_ rightarrow ... rightarrow S_kS'. By bf claim LSLS_LS_...LS_kLS'.
Contained in these collections:

