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EleoNora [17]
3 years ago
11

Given the following matrices A and B, find an invertible matrix U such that UA = B:

Mathematics
1 answer:
Naddik [55]3 years ago
3 0

<span>A and B must be invertible, we have UA=B, since A is invertible.  A^-1 exists, by multiplying with A^-1, we have UA A^-1 =B A^-1. But AA^-1 = I (identity matrix) and XI=X, for all matrix X, we find UI= B A^-1, and U= B A^-1.</span>

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Twice the first number is eleven more than the sum of the other two numbers. The sum of twice the first number and three times t
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first sentence says twice first number so 2x. is means equals so =. eleven more than sum of the other two numbers. sum of other two numbers is (y + z) and eleven more than that is (y + z) + 11. So so this sentence says:
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second sentence says sum of twice the first and three times third. twice first is 2x and three times third is 3z. their sum would be (2x + 3z). It it says is one more than second number. So = (y + 1). So so this means:
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now for the work.

We we can easily solve for a by subtracting the first 2 equations. the way to subtract 2 equations is by subtracting one left side of equal sign from the other equations left side and then doing the same with the right side.

So we will subtract the 1st and second equations we made.
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now we can plug in z into the equations and subtract second and third equations. But but we will subtract opposite sides. So so left minus right and right mi is left after we plug in z:
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other one would be
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1

So so put an equal sign and get:
x - 5 = 1
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5 0
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Answer:

\huge\boxed{9}

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