$A=pmatrix1&2&3&4&5\2&3&4&5&6.$ Find $det(A^TA)$. [duplicate]
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When A and B are of different order given the $det(AB)$,then calculate $det(BA)$
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Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.
I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?
linear-algebra matrices determinant
marked as duplicate by amd, amWhy
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Aug 19 at 22:55
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This question already has an answer here:
When A and B are of different order given the $det(AB)$,then calculate $det(BA)$
3 answers
Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.
I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?
linear-algebra matrices determinant
marked as duplicate by amd, amWhy
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Aug 19 at 22:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Related : math.stackexchange.com/questions/1782600/â¦
â Arnaud D.
Aug 19 at 15:10
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up vote
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up vote
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down vote
favorite
This question already has an answer here:
When A and B are of different order given the $det(AB)$,then calculate $det(BA)$
3 answers
Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.
I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?
linear-algebra matrices determinant
This question already has an answer here:
When A and B are of different order given the $det(AB)$,then calculate $det(BA)$
3 answers
Suppose $$A=pmatrix1&2&3&4&5\2&3&4&5&6$$ Find $det(A^TA)$.
I know exactly how to calculate it by writing it as a $5times5$ matrix. But how to calculate it smartly?
This question already has an answer here:
When A and B are of different order given the $det(AB)$,then calculate $det(BA)$
3 answers
linear-algebra matrices determinant
linear-algebra matrices determinant
edited Aug 19 at 22:54
amWhy
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190k27221433
asked Aug 19 at 14:45
LOIS
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marked as duplicate by amd, amWhy
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Related : math.stackexchange.com/questions/1782600/â¦
â Arnaud D.
Aug 19 at 15:10
add a comment |Â
Related : math.stackexchange.com/questions/1782600/â¦
â Arnaud D.
Aug 19 at 15:10
Related : math.stackexchange.com/questions/1782600/â¦
â Arnaud D.
Aug 19 at 15:10
Related : math.stackexchange.com/questions/1782600/â¦
â Arnaud D.
Aug 19 at 15:10
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1 Answer
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Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.
Hence the determinant must be $0$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
30
down vote
accepted
Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.
Hence the determinant must be $0$.
add a comment |Â
up vote
30
down vote
accepted
Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.
Hence the determinant must be $0$.
add a comment |Â
up vote
30
down vote
accepted
up vote
30
down vote
accepted
Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.
Hence the determinant must be $0$.
Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.
Hence the determinant must be $0$.
answered Aug 19 at 14:49
Siong Thye Goh
82.7k1456104
82.7k1456104
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Related : math.stackexchange.com/questions/1782600/â¦
â Arnaud D.
Aug 19 at 15:10