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Fiesta28 [93]
3 years ago
9

Show that if A is​ invertible, then det Upper A Superscript negative 1det A−1equals=StartFraction 1 Over det Upper A EndFraction

1 det A. What​ theorem(s) should be used to examine the quantity det Upper A Superscript negative 1det A−1​? Select all that apply. A. If A and B are ntimes×n ​matrices, then det ABequals=(det Upper A )(det A)(det Upper B )(det B). Your answer is correct.B. If one row of a square matrix A is multiplied by k to produce​ B, then det Upper Bdet Bequals=ktimes•(det Upper A )(det A). C. A square matrix A is invertible if and only if det Upper Adet Anot equals≠0. Your answer is correct.D. If A is an ntimes×n ​matrix, then det Upper A Superscript Upper Tdet ATequals=det Upper Adet A. Consider the quantity (det Upper A )(det Upper A Superscript negative 1 Baseline )(det A)det A−1. To what must this be​ equal? A. det Upper I
Mathematics
1 answer:
netineya [11]3 years ago
7 0

Answer:

Therefore  

 det(A)*det(A^{-1}) = 1      and         det(A^{-1}) = 1/det(A)

Step-by-step explanation:

Remember that  if  

I  =  Identity Matrix    

then    det(I)=1.

Also remember that for any invertible matrix  

A*A^{-1}  = I =Identity Matrix.

Therefore    

det(A*A^{-1})  = det(A)*det(A^{-1}) = det(I) = 1

Therefore  

 det(A)*det(A^{-1}) = 1      and         det(A^{-1}) = 1/det(A)

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Answer:

The number of deserters is 34.

Step-by-step explanation:

We have to calculate the number of desertors in a group of 1500 soldiers.

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If he divide in groups of 7, there are three left over. If we take 3, the number of soldiers gruoped in 7 has to end in 8 or 3. The only numbers bigger than 1400 that end in 8 or 3 and have 7 as common divider are 1428 and 1463.

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