Determinant and inverse 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
Solution
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Exercise:
A is invertible iff ad-bcneq . Moreover if A is invertible then A^-fracad-bc pmatrix d & -b -c & a pmatrix. The number ad-bc is textdetA the determinant of A.
Solution:
Proof. A textis not invertible &iff textrankA &iff textits columns are linearly depent &iff leftarrayc a c arrayright lambda leftarrayc b d arrayright textor leftarrayc b d arrayright tau leftarrayc a c arrayright textIf leftarrayc a c arrayright lambda leftarrayc b d arrayright &Longrightarrow alambda b clambda d Longrightarrow ad-bc lambda bd- lambda bd textIf leftarrayc b d arrayright tau leftarrayc a c arrayright &Longrightarrow btau a dtau c Longrightarrow ad-bc tau ac-tau ac We have proven that A is not invertible Longrightarrow ad-bc. Conversely ase ad-bc. textIf a&Longrightarrow bc textso either b textor c &Longrightarrow texteither A pmatrix & c & d pmatrix textor A pmatrix & b & d pmatrix. and in both cases A is not invertible. textSo ase aneq &Longrightarrow dfracbca &Longrightarrow A pmatrix a & b c & fracbca pmatrix &Longrightarrow textthe second col of A is a multiple of the first col. &Longrightarrow textindeed leftarrayc b fracbca arrayright fracba leftarrayc a c arrayright &Longrightarrow textthe columns of A are linearly depent &Longrightarrow A textis not invertible. A is not invertible iff ad-bc. The formula for A^- direct calculation. fracad-bc pmatrix d & -b -c & a pmatrix pmatrix a & b c & d pmatrix fracad-bc pmatrix ad-bc & bd-bd -ac+ac & -bc+ad pmatrix fracad-bc pmatrix ad-bc & & -bc+ad pmatrix pmatrix & & pmatrix
A is invertible iff ad-bcneq . Moreover if A is invertible then A^-fracad-bc pmatrix d & -b -c & a pmatrix. The number ad-bc is textdetA the determinant of A.
Solution:
Proof. A textis not invertible &iff textrankA &iff textits columns are linearly depent &iff leftarrayc a c arrayright lambda leftarrayc b d arrayright textor leftarrayc b d arrayright tau leftarrayc a c arrayright textIf leftarrayc a c arrayright lambda leftarrayc b d arrayright &Longrightarrow alambda b clambda d Longrightarrow ad-bc lambda bd- lambda bd textIf leftarrayc b d arrayright tau leftarrayc a c arrayright &Longrightarrow btau a dtau c Longrightarrow ad-bc tau ac-tau ac We have proven that A is not invertible Longrightarrow ad-bc. Conversely ase ad-bc. textIf a&Longrightarrow bc textso either b textor c &Longrightarrow texteither A pmatrix & c & d pmatrix textor A pmatrix & b & d pmatrix. and in both cases A is not invertible. textSo ase aneq &Longrightarrow dfracbca &Longrightarrow A pmatrix a & b c & fracbca pmatrix &Longrightarrow textthe second col of A is a multiple of the first col. &Longrightarrow textindeed leftarrayc b fracbca arrayright fracba leftarrayc a c arrayright &Longrightarrow textthe columns of A are linearly depent &Longrightarrow A textis not invertible. A is not invertible iff ad-bc. The formula for A^- direct calculation. fracad-bc pmatrix d & -b -c & a pmatrix pmatrix a & b c & d pmatrix fracad-bc pmatrix ad-bc & bd-bd -ac+ac & -bc+ad pmatrix fracad-bc pmatrix ad-bc & & -bc+ad pmatrix pmatrix & & pmatrix
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Exercise:
A is invertible iff ad-bcneq . Moreover if A is invertible then A^-fracad-bc pmatrix d & -b -c & a pmatrix. The number ad-bc is textdetA the determinant of A.
Solution:
Proof. A textis not invertible &iff textrankA &iff textits columns are linearly depent &iff leftarrayc a c arrayright lambda leftarrayc b d arrayright textor leftarrayc b d arrayright tau leftarrayc a c arrayright textIf leftarrayc a c arrayright lambda leftarrayc b d arrayright &Longrightarrow alambda b clambda d Longrightarrow ad-bc lambda bd- lambda bd textIf leftarrayc b d arrayright tau leftarrayc a c arrayright &Longrightarrow btau a dtau c Longrightarrow ad-bc tau ac-tau ac We have proven that A is not invertible Longrightarrow ad-bc. Conversely ase ad-bc. textIf a&Longrightarrow bc textso either b textor c &Longrightarrow texteither A pmatrix & c & d pmatrix textor A pmatrix & b & d pmatrix. and in both cases A is not invertible. textSo ase aneq &Longrightarrow dfracbca &Longrightarrow A pmatrix a & b c & fracbca pmatrix &Longrightarrow textthe second col of A is a multiple of the first col. &Longrightarrow textindeed leftarrayc b fracbca arrayright fracba leftarrayc a c arrayright &Longrightarrow textthe columns of A are linearly depent &Longrightarrow A textis not invertible. A is not invertible iff ad-bc. The formula for A^- direct calculation. fracad-bc pmatrix d & -b -c & a pmatrix pmatrix a & b c & d pmatrix fracad-bc pmatrix ad-bc & bd-bd -ac+ac & -bc+ad pmatrix fracad-bc pmatrix ad-bc & & -bc+ad pmatrix pmatrix & & pmatrix
A is invertible iff ad-bcneq . Moreover if A is invertible then A^-fracad-bc pmatrix d & -b -c & a pmatrix. The number ad-bc is textdetA the determinant of A.
Solution:
Proof. A textis not invertible &iff textrankA &iff textits columns are linearly depent &iff leftarrayc a c arrayright lambda leftarrayc b d arrayright textor leftarrayc b d arrayright tau leftarrayc a c arrayright textIf leftarrayc a c arrayright lambda leftarrayc b d arrayright &Longrightarrow alambda b clambda d Longrightarrow ad-bc lambda bd- lambda bd textIf leftarrayc b d arrayright tau leftarrayc a c arrayright &Longrightarrow btau a dtau c Longrightarrow ad-bc tau ac-tau ac We have proven that A is not invertible Longrightarrow ad-bc. Conversely ase ad-bc. textIf a&Longrightarrow bc textso either b textor c &Longrightarrow texteither A pmatrix & c & d pmatrix textor A pmatrix & b & d pmatrix. and in both cases A is not invertible. textSo ase aneq &Longrightarrow dfracbca &Longrightarrow A pmatrix a & b c & fracbca pmatrix &Longrightarrow textthe second col of A is a multiple of the first col. &Longrightarrow textindeed leftarrayc b fracbca arrayright fracba leftarrayc a c arrayright &Longrightarrow textthe columns of A are linearly depent &Longrightarrow A textis not invertible. A is not invertible iff ad-bc. The formula for A^- direct calculation. fracad-bc pmatrix d & -b -c & a pmatrix pmatrix a & b c & d pmatrix fracad-bc pmatrix ad-bc & bd-bd -ac+ac & -bc+ad pmatrix fracad-bc pmatrix ad-bc & & -bc+ad pmatrix pmatrix & & pmatrix
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