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notsponge [240]
2 years ago
9

Is the statement "Two matrices are row equivalent if they have the same number of rows" true or false? Explain. A. True, because

two matrices that are row equivalent have the same number of solutions, which means that they have the same number of rows. B. True, because two matrices are row equivalent if they have the same number of rows and column equivalent if they have the same number of columns. C. False, because if two matrices are row equivalent it means that there exists a sequence of row operations that transforms one matrix to the other. D. False, because if two matrices are row equivalent it means that they have the same number of row solutions.
Mathematics
1 answer:
vlabodo [156]2 years ago
7 0

Answer: The answer is (C).


Step-by-step explanation: The given statement is - "Two matrices are row equivalent if they have the same number of rows". We are to explain whether the statement is true or false.

What are row equivalent matrices? The answer to this question is -

Two matrices are said to be row equivalent if one of the matrices can be obtained from the other by applying a number of elementary row operations. Or, we can say two matrices of same order are row equivalent if they have same row space.

Thus, the correct option is (C).



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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

Area = 128

Let the dimension of the paper be x and y;

Such that:

Length = x

Width = y

So:

Area = x * y

Substitute 128 for Area

128 = x * y

Make x the subject

x = \frac{128}{y}

When 1 inch margin is at top and bottom

The length becomes:

Length = x + 1 + 1

Length = x + 2

When 2 inch margin is at both sides

The width becomes:

Width = y + 2 + 2

Width = y + 4

The New Area (A) is then calculated as:

A = (x + 2) * (y + 4)

Substitute \frac{128}{y} for x

A = (\frac{128}{y} + 2) * (y + 4)

Open Brackets

A = 128 + \frac{512}{y} + 2y + 8

Collect Like Terms

A = \frac{512}{y} + 2y + 8+128

A = \frac{512}{y} + 2y + 136

A= 512y^{-1} + 2y + 136

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

A' = -512y^{-2} + 2

Set

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Collect Like Terms

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512 = 2y^2

Divide through by 2

256=y^2

Take square roots of both sides

\sqrt{256=y^2

16=y

y = 16

Recall that:

x = \frac{128}{y}

x = \frac{128}{16}

x = 8

Recall that the new dimensions are:

Length = x + 2

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So:

Length = 8 + 2

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Width = 16 + 4

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To double-check;

Differentiate A'

A' = -512y^{-2} + 2

A" = -2 * -512y^{-3}

A" = 1024y^{-3}

A" = \frac{1024}{y^3}

The above value is:

A" = \frac{1024}{y^3} > 0

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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Answer:

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Answer:

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