2 years ago

The diagram below shows rectangle ABC is a midtsin of

Mathematics

1 answer:

Oduvanchick [21]2 years ago

8 0

Answer:

4x2+3=

Step-by-step explanation:

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Ne4ueva [31]

Answer:

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Step-by-step explanation:

=10 3/5 + 2 3/5

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2 years ago

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2 years ago

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

netineya [11]

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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3 years ago

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Ludmilka [50]

Answer:

17.76

Step-by-step explanation:

0.44 x 4 = 1.76

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3 years ago