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MaRussiya [10]
3 years ago
5

Show that det(A) = 0 without directly evaluating the determinant. A = [-4 1 1 1 1; 1 -4 1 1 1; 1 1 -4 1 1; 1 1 1 -4 1; 1 1 1 1 -

4]
Mathematics
2 answers:
Anuta_ua [19.1K]3 years ago
8 0
You could interpret them as system of equation and answer the question easily

from A 

-4x + y + x + w =1

write in same form from A 

as this system has five variables and four of them are unknowns it means the unique solution is not possible 

So, det (A) = 0



Allisa [31]3 years ago
5 0

Answer:You could interpret them as system of equation and answer the question easily

from A 

-4x + y + x + w =1

write in same form from A 

as this system has five variables and four of them are unknowns it means the unique solution is not possible 

So, det (A) = 0

Read more on Brainly.com - brainly.com/question/2253943#readmore

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Evaluate the expression below when x = 4 and y = 12.<br> 5х^2 — Зу
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Answer:

44

Step-by-step explanation:

x = 4 and y = 12.

5х² - 3у

20x - 3y

80 - 36

44

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3 years ago
A building that is 100 ft tall casts a shadow that makes a 30 degree angle with the ground. Approximately how long, in feet, is
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Read 2 more answers
13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult
Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that \frac{ab}{m} is and integer.

As m= vb, then \frac{m}{b} is an integer, lets say \frac{m}{b}=v; and as m=ua, then \frac{m}{a}=u.

So \frac{ab}{m}v=\frac{ab}{m}\frac{m}{b}=a, so \frac{ab}{m} divides a; on the other hand, \frac{ab}{m}u=\frac{ab}{m}\frac{m}{a}=b, so \frac{ab}{m} divides b. From this we can conclude that \frac{ab}{m} is a common divisor of a and b.

4 0
3 years ago
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Likurg_2 [28]

Answer:

2/1and6/3 because if you do cross multilication they both equal 6

Step-by-step explanation:

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3 years ago
How do you solve the literal equation y=4/5x-9?
Wewaii [24]

Answer:

todays my birthday

Step-by-step explanation:

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