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Luda [366]
3 years ago
13

Let A and B be square matrices. Show that if ABequals=I then A and B are both​ invertible, with Bequals=Upper A Superscript nega

tive 1A−1 and Aequals=Upper B Superscript negative 1B−1.
Mathematics
1 answer:
stich3 [128]3 years ago
5 0

Step-by-step explanation:

Since AB=I, we have

det(A)det(B)=det(AB)=det(I)=1.

This implies that the determinants det(A) and det(B) are not zero.

Hence A,B are invertible matrices: A−1,B−1 exist.

Now we compute

I=BB−1=BIB−1=B(AB)B−1=BAI=BA.since AB=I

Hence we obtain BA=I.

Since AB=I and BA=I, we conclude that B=A−1.

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<em />

So, the number of different arrangement is

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Substitute known values

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^7P_7 = \frac{7!}{0!}

^7P_7 = \frac{5040}{1}

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Read more about arrangements at:

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Which statement is true about the end behavior of the
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As the x-values go to positive infinity, the function's

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