Invertible matrices
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
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Exercise:
Let Ain M_ntimes nK. Place A together with I_n in the following way A|I_n. Then A is invertible iff A|I_n can be brought via a sequence of elementary row-operations to the form I_n|B for some matrix Bin M_ntimes nK. In this case we have A^-B.
Solution:
bf Lemma. Let BCDin M_ntimes nK. Then B C|DB C|B D. Proof of Lemma. Follows directly from the definition of matrix multiplication. Proof of the theorem. Ase A is invertible. By the previous theorem exists elementary matrices T_...T_k s.t. T_k T_k- ... T_ AI_n. Longrightarrow T_k T_k- ... T_ A|I_nI_n|T_k T_k- ... T_. We also have: A B T_^-... T_k-^- T_k^- T_k T_k-... T_...I_n B A T_k^- T_k-^- ... T_ T_^-... T_k-^- T_k^-...I_n &Longrightarrow A^-B Conversely ase that A|I_n can be brought to I_n|B by a sequence of elementary row-operations for some matrix Bin M_ntimes nK Longrightarrow A itself is row-equivalent to I_n Longrightarrow A has rank n Longrightarrow A is invertible.
Let Ain M_ntimes nK. Place A together with I_n in the following way A|I_n. Then A is invertible iff A|I_n can be brought via a sequence of elementary row-operations to the form I_n|B for some matrix Bin M_ntimes nK. In this case we have A^-B.
Solution:
bf Lemma. Let BCDin M_ntimes nK. Then B C|DB C|B D. Proof of Lemma. Follows directly from the definition of matrix multiplication. Proof of the theorem. Ase A is invertible. By the previous theorem exists elementary matrices T_...T_k s.t. T_k T_k- ... T_ AI_n. Longrightarrow T_k T_k- ... T_ A|I_nI_n|T_k T_k- ... T_. We also have: A B T_^-... T_k-^- T_k^- T_k T_k-... T_...I_n B A T_k^- T_k-^- ... T_ T_^-... T_k-^- T_k^-...I_n &Longrightarrow A^-B Conversely ase that A|I_n can be brought to I_n|B by a sequence of elementary row-operations for some matrix Bin M_ntimes nK Longrightarrow A itself is row-equivalent to I_n Longrightarrow A has rank n Longrightarrow A is invertible.
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Exercise:
Let Ain M_ntimes nK. Place A together with I_n in the following way A|I_n. Then A is invertible iff A|I_n can be brought via a sequence of elementary row-operations to the form I_n|B for some matrix Bin M_ntimes nK. In this case we have A^-B.
Solution:
bf Lemma. Let BCDin M_ntimes nK. Then B C|DB C|B D. Proof of Lemma. Follows directly from the definition of matrix multiplication. Proof of the theorem. Ase A is invertible. By the previous theorem exists elementary matrices T_...T_k s.t. T_k T_k- ... T_ AI_n. Longrightarrow T_k T_k- ... T_ A|I_nI_n|T_k T_k- ... T_. We also have: A B T_^-... T_k-^- T_k^- T_k T_k-... T_...I_n B A T_k^- T_k-^- ... T_ T_^-... T_k-^- T_k^-...I_n &Longrightarrow A^-B Conversely ase that A|I_n can be brought to I_n|B by a sequence of elementary row-operations for some matrix Bin M_ntimes nK Longrightarrow A itself is row-equivalent to I_n Longrightarrow A has rank n Longrightarrow A is invertible.
Let Ain M_ntimes nK. Place A together with I_n in the following way A|I_n. Then A is invertible iff A|I_n can be brought via a sequence of elementary row-operations to the form I_n|B for some matrix Bin M_ntimes nK. In this case we have A^-B.
Solution:
bf Lemma. Let BCDin M_ntimes nK. Then B C|DB C|B D. Proof of Lemma. Follows directly from the definition of matrix multiplication. Proof of the theorem. Ase A is invertible. By the previous theorem exists elementary matrices T_...T_k s.t. T_k T_k- ... T_ AI_n. Longrightarrow T_k T_k- ... T_ A|I_nI_n|T_k T_k- ... T_. We also have: A B T_^-... T_k-^- T_k^- T_k T_k-... T_...I_n B A T_k^- T_k-^- ... T_ T_^-... T_k-^- T_k^-...I_n &Longrightarrow A^-B Conversely ase that A|I_n can be brought to I_n|B by a sequence of elementary row-operations for some matrix Bin M_ntimes nK Longrightarrow A itself is row-equivalent to I_n Longrightarrow A has rank n Longrightarrow A is invertible.
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