Determinants of block 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:
Consider a block matrix M of the type M leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright where Ain M_rtimes rK Cin M_stimes sK Bin M_rtimes sK. Proof that textdetMtextdetA textdetC.
Solution:
Proof. Define DABC:textdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright as a function D:M_rtimes rKtimes M_rtimes sKtimes M_stimes sK longrightarrow K. If we fix the first two matrices A and B then the function M_stimes sK longrightarrow K Clongmapsto DABC is s-linear and alternating. By a previous theorem we have DABCtextdetC DABI_s. Let's calculate DABI_s. By definition DABI_s leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright By doing row operations of the type R_i+cR_jlongrightarrow R_k jneq i cin K we can use the last s-rows of leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright to get to the matrix leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. By a previous theorem det leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright textdetleftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. The function M_rtimes rKlongrightarrow K Alongmapsto det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright is r-linear and alternating. By a previous result we have det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright textdetAtextdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright textdetA textdetI_r+s textdetA going back to the results above we get that det leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright DABC:textdetA textdetC.
Consider a block matrix M of the type M leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright where Ain M_rtimes rK Cin M_stimes sK Bin M_rtimes sK. Proof that textdetMtextdetA textdetC.
Solution:
Proof. Define DABC:textdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright as a function D:M_rtimes rKtimes M_rtimes sKtimes M_stimes sK longrightarrow K. If we fix the first two matrices A and B then the function M_stimes sK longrightarrow K Clongmapsto DABC is s-linear and alternating. By a previous theorem we have DABCtextdetC DABI_s. Let's calculate DABI_s. By definition DABI_s leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright By doing row operations of the type R_i+cR_jlongrightarrow R_k jneq i cin K we can use the last s-rows of leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright to get to the matrix leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. By a previous theorem det leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright textdetleftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. The function M_rtimes rKlongrightarrow K Alongmapsto det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright is r-linear and alternating. By a previous result we have det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright textdetAtextdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright textdetA textdetI_r+s textdetA going back to the results above we get that det leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright DABC:textdetA textdetC.
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Exercise:
Consider a block matrix M of the type M leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright where Ain M_rtimes rK Cin M_stimes sK Bin M_rtimes sK. Proof that textdetMtextdetA textdetC.
Solution:
Proof. Define DABC:textdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright as a function D:M_rtimes rKtimes M_rtimes sKtimes M_stimes sK longrightarrow K. If we fix the first two matrices A and B then the function M_stimes sK longrightarrow K Clongmapsto DABC is s-linear and alternating. By a previous theorem we have DABCtextdetC DABI_s. Let's calculate DABI_s. By definition DABI_s leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright By doing row operations of the type R_i+cR_jlongrightarrow R_k jneq i cin K we can use the last s-rows of leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright to get to the matrix leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. By a previous theorem det leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright textdetleftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. The function M_rtimes rKlongrightarrow K Alongmapsto det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright is r-linear and alternating. By a previous result we have det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright textdetAtextdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright textdetA textdetI_r+s textdetA going back to the results above we get that det leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright DABC:textdetA textdetC.
Consider a block matrix M of the type M leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright where Ain M_rtimes rK Cin M_stimes sK Bin M_rtimes sK. Proof that textdetMtextdetA textdetC.
Solution:
Proof. Define DABC:textdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright as a function D:M_rtimes rKtimes M_rtimes sKtimes M_stimes sK longrightarrow K. If we fix the first two matrices A and B then the function M_stimes sK longrightarrow K Clongmapsto DABC is s-linear and alternating. By a previous theorem we have DABCtextdetC DABI_s. Let's calculate DABI_s. By definition DABI_s leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright By doing row operations of the type R_i+cR_jlongrightarrow R_k jneq i cin K we can use the last s-rows of leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright to get to the matrix leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. By a previous theorem det leftarray@c|c@ matrix A matrix & B hline matrix matrix & I_s arrayright textdetleftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright. The function M_rtimes rKlongrightarrow K Alongmapsto det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright is r-linear and alternating. By a previous result we have det leftarray@c|c@ matrix A matrix & hline matrix matrix & I_s arrayright textdetAtextdet leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright textdetA textdetI_r+s textdetA going back to the results above we get that det leftarray@c|c@ matrix A matrix & B hline matrix matrix & C arrayright DABC:textdetA textdetC.
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