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Vera_Pavlovna [14]
3 years ago
7

Is the statement​ "Elementary row operations on an augmented matrix never change the solution set of the associated linear​ syst

em" true or​ false? Explain.
Mathematics
1 answer:
julia-pushkina [17]3 years ago
7 0

Answer:

True

Step-by-step explanation:

A matrix is a rectangular array in which elements are arranged in rows and columns.

An augmented matrix is a matrix in which same row operations are performed on both the sides of equal signs in the given linear system of equations.

Elementary row operations are the operations like multiplication or division which are performed in the original matrix to get the elementary matrix.

True, elementary row operations on an augmented matrix never change the solution set of the associated linear​ system as the elementary row operations replace a linear system with an equivalent linear system.

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<span>280 I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself. You can choose 1 of 7 appetizers. So we have n = 7 After that, you chose an entre, so the number of possible meals to this point is n = 7 * 10 = 70 Finally, you finish off with a dessert, so the number of meals is: n = 70 * 4 = 280 Therefore the number of possible meals you can have is 280. Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is n = 77 * 1010 * 44 = 3421880 But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
5 0
3 years ago
Show the working out too please in numbers, thank you.
vazorg [7]

Hii :))

\tt \: - 5 =  \frac{x}{6}  \\   \tt \:  - 5 \times 6 = x \\  \boxed{  \tt \:  - 30 = x}

__________________

  • The correct value of x is <u>-</u><u> </u><u>3</u><u>0</u><u>.</u>

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