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Sidana [21]
3 years ago
12

Pls help me with this question :)

Mathematics
1 answer:
natima [27]3 years ago
3 0
I think it would be 4
:)
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The Venn diagram shows all of the elements in sets A, B and ξ. Find the elements in (A ∩ B) U (A U B)’
marusya05 [52]

This is impossible to answer without knowing what the sets A and B contain (and what ξ even refers to - universal set?).

However, we have

(A U B)' = A' ∩ B'

so that

(A ∩ B) U (A U B)' = (A ∩ B) U (A' ∩ B') = (A U A') ∩ (B U B')

If ξ is indeed the universal set, then both A U A' = ξ and B U B' = ξ, so we end up with ξ ∩ ξ = ξ.

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2 years ago
A teacher asked her students how many pets they own. Here
AlekseyPX

Step-by-step explanation:

the answer is (c)

because 0 is 3 in number,

1 is 4 in number,

2 is 3in number

3 is 2 in number,

4 is 2 in number,

no 5,6,7

8 is 1 in number

and no 9

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2 years ago
What is the sum of -5 and -5?<br> -10<br> 10<br> 0
Kipish [7]

Answer:

-10

Step-by-step explanation:

5 0
3 years ago
Bill and 10 friends are planning for a pizza party.They purchased 3 quarts of juice.If each glass holds 8 ounces will everyone g
amm1812
Yes. all the friends will get at least one cup of juice
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3 years ago
Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the
Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right], to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B=\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]

the system Bx=0 can be represented in matrix form as:

\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right] ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have B_{0}:      

B_{0}=\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]. The rank of B_{0} can be found by using the second column and third column pair as follows:

|B_{0}|=(3*0)-(0*2)=0 i.e, B_{0} is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is \neq0.

Comparing the rank of both B and B_{0}, it is obvious that

Rank of B\neqRank of B_{0} since (-2)<1.

Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution.     </em>

(2) If B=\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right] is the transpose of matrix A=\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right], then

Then the equation Bx=0 is represented as:

\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right]..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0  i.e B has a rank of order 1.

B_{0}=\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right],

|B_{0}|=(5*0)-(0*10)=0-0=0   i.e B_{0} has a rank of order 1.

we can therefor conclude that since

rank B=rank B_{0}=1,  equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (X_{1} and X_{2}).

<u>Summary:</u>

  • Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
  • Determine the rank of both the coefficients matrix B and B_{0} which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
  • If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.
5 0
3 years ago
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