(a - ab) / a² ÷ a - 1 / a³

Image below. :)

Explanation:

Step by step explanation along with Matlab code and output is provided below.

Step-by-step explanation:

We are given three matrices A, B, and C of size 2x2

A = [0 1; 0 0]

B =[1 2; -3 -6]

C =[4 -2; -2 1]

Output:

A =  0     1

     0     0

B =  1     2

    -3    -6

C =   4    -2

    -2     1

Let us first check if the given matrices A, B, and C are singular or not

% the Matlab function det( ) calculates the determinant of a matrix

det_A=det(A)

det_B=det(B)

det_C=det(C)

Output:

det_A =  0

det_B =  3.3307e-16  (its practically zero)

det_C =  0

So the given matrices are singular which means that the determinant of the matrix is zero so inverse of these matrices is not possible.

Rule I:

a1=B*C

Output:

a1 =  0     0

      0     0

In Matrix theory, if BC=0 then B=0 or C=0 doesn't hold true

For matrices B*C=0 does not imply that either B or C is zero matrix but rather it implies that at least one of them is singular. In this case we know that both B and C are singular matrices therefore, BC=0  

Rule II:

a2=A^2

Output:

a2=  0     0

      0     0

In Matrix theory, if A^2=0 then A=0 doesn't hold true

For matrices A^2=0 does not imply that A is zero matrix but rather it implies that A is singular. We already know that A is singular therefore, A^2=0

Rule III:

a3_L=(A+B)^2

a3_R=A^2+2*A*B+B^2

Output:

a3_L =    -8   -15

            15    27

a3_R =   -11   -22

             15    30

In Matrix theory, (A + B)^2 = A^2 + 2AB + B^2 doesn't hold true.

(A + B)^2 = A^2 + 2AB + B^2 might hold true if AB = BA,  but generally, AB≠BA in matrix algebra.

Rule IV:  

a4_L=(A-B)*(A+B)

a4_R=A^2-B^2

Output:

a4_L =     2     3

            -15   -27

a4_R =    5    10

            -15   -30

In Matrix theory, (A-B)(A+B)  = A^2-B^2  doesn't hold true.

Rule V:    

a5_L=A*(B+C)

a5_R=A*B+A*C

Output:

a5_L =    -5    -5

              0     0

a5_R =   -5    -5

              0     0

In Matrix theory, A(B+C) =  AB+AC  holds true.

Rule VI:    

a6_L=A*(B+C)

a6_R=B*A+C*A

Output:

a6_L =    -5    -5

              0     0

a6_R =    0     5

              0    -5

In Matrix theory, A(B+C) =  BA+CA  doesn't hold true.

on a side note;  (B+C)A =  BA+CA holds true

Rule VII:  

a7_L=(A*B)^2

a7_R=A^2*B^2

Output:

a7_L =     9    18

              0     0

a7_R =     0     0

              0     0

In Matrix theory, (AB)^2  =A^2*B^2   doesn't hold true.

(AB)^2  =A^2*B^2   might hold true if and only if  BA=AB  which is not true in general.  

Step by step explanation along with Matlab code and output is provided below.

Step-by-step explanation:

We are given three matrices A, B, and C of size 2x2

A = [0 1; 0 0]

B =[1 2; -3 -6]

C =[4 -2; -2 1]

Output:

A =  0     1

     0     0

B =  1     2

    -3    -6

C =   4    -2

    -2     1

Let us first check if the given matrices A, B, and C are singular or not

% the Matlab function det( ) calculates the determinant of a matrix

det_A=det(A)

det_B=det(B)

det_C=det(C)

Output:

det_A =  0

det_B =  3.3307e-16  (its practically zero)

det_C =  0

So the given matrices are singular which means that the determinant of the matrix is zero so inverse of these matrices is not possible.

Rule I:

a1=B*C

Output:

a1 =  0     0

      0     0

In Matrix theory, if BC=0 then B=0 or C=0 doesn't hold true

For matrices B*C=0 does not imply that either B or C is zero matrix but rather it implies that at least one of them is singular. In this case we know that both B and C are singular matrices therefore, BC=0  

Rule II:

a2=A^2

Output:

a2=  0     0

      0     0

In Matrix theory, if A^2=0 then A=0 doesn't hold true

For matrices A^2=0 does not imply that A is zero matrix but rather it implies that A is singular. We already know that A is singular therefore, A^2=0

Rule III:

a3_L=(A+B)^2

a3_R=A^2+2*A*B+B^2

Output:

a3_L =    -8   -15

            15    27

a3_R =   -11   -22

             15    30

In Matrix theory, (A + B)^2 = A^2 + 2AB + B^2 doesn't hold true.

(A + B)^2 = A^2 + 2AB + B^2 might hold true if AB = BA,  but generally, AB≠BA in matrix algebra.

Rule IV:  

a4_L=(A-B)*(A+B)

a4_R=A^2-B^2

Output:

a4_L =     2     3

            -15   -27

a4_R =    5    10

            -15   -30

In Matrix theory, (A-B)(A+B)  = A^2-B^2  doesn't hold true.

Rule V:    

a5_L=A*(B+C)

a5_R=A*B+A*C

Output:

a5_L =    -5    -5

              0     0

a5_R =   -5    -5

              0     0

In Matrix theory, A(B+C) =  AB+AC  holds true.

Rule VI:    

a6_L=A*(B+C)

a6_R=B*A+C*A

Output:

a6_L =    -5    -5

              0     0

a6_R =    0     5

              0    -5

In Matrix theory, A(B+C) =  BA+CA  doesn't hold true.

on a side note;  (B+C)A =  BA+CA holds true

Rule VII:  

a7_L=(A*B)^2

a7_R=A^2*B^2

Output:

a7_L =     9    18

              0     0

a7_R =     0     0

              0     0

In Matrix theory, (AB)^2  =A^2*B^2   doesn't hold true.

(AB)^2  =A^2*B^2   might hold true if and only if  BA=AB  which is not true in general.  

True:

Step-by-step explanation:

1.

Correct property:

The given property is not true.

2.

Correct property:

the given property is not true.

3.

The given property is true.

4.

The given property is true.

Property:

True:

Step-by-step explanation:

1.

Correct property:

The given property is not true.

2.

Correct property:

the given property is not true.

3.

The given property is true.

4.

The given property is true.

Property:

i) P(A) ; The probability that event A occurs

ii) P(AB) ; The probability that both event A  and event B occur

iii) P(AUB) ; The probability that either event A  or event B occurs

iv) 1 - P(ANB) ; The probability that both events A  and B do not occur                                 together, but either  may occur by itself

v) 1- P(AUB) ; The probability that neither event A  or event B occurs

vi) P(AIB) ; The probability that event A occurs  given the fact that event B occurs

Step-by-step explanation:

i)

P(A) simply represents the probability that an event A will occur. This event could be passing an examination, having snow in summer, arriving to work on time and so forth.

ii)

P(AB) is simply the probability that both event A  and event B do occur. This is usually given by the product of the individual probabilities. Event A could be rolling a 6 in one throw of a fair die while B could be the event that a fair coin lands heads in a single toss.

iii)

P(AUB) refers to the probability that either event A  or event B occurs. This is read out as the probability of A union B. This is usually given by the sum of the individual probabilities.

iv)

1 - P(ANB) is the probability that both events A  and B do not occur                                 together, but either  may occur by itself. P(ANB) is the probability that both events A  and B occur together. This is read out as the probability of A intersection B. Therefore implying that 1 - P(ANB) is simply the probability that either event A or B occurs but A and B can not occur together.

v)

1- P(AUB) refers to the probability that neither event A  or event B occurs. Earlier we defined P(AUB) as the probability that either event A  or event B occurs. 1- P(AUB) simply the complement of P(AUB).

vi)

P(AIB) refers to the probability that event A occurs  given the fact that event B occurs. This is a conditional probability event which evaluates the likelihood of an event A occurring given that an associated event B has already occurred

i) P(A) ; The probability that event A occurs

ii) P(AB) ; The probability that both event A  and event B occur

iii) P(AUB) ; The probability that either event A  or event B occurs

iv) 1 - P(ANB) ; The probability that both events A  and B do not occur                                 together, but either  may occur by itself

v) 1- P(AUB) ; The probability that neither event A  or event B occurs

vi) P(AIB) ; The probability that event A occurs  given the fact that event B occurs

Step-by-step explanation:

i)

P(A) simply represents the probability that an event A will occur. This event could be passing an examination, having snow in summer, arriving to work on time and so forth.

ii)

P(AB) is simply the probability that both event A  and event B do occur. This is usually given by the product of the individual probabilities. Event A could be rolling a 6 in one throw of a fair die while B could be the event that a fair coin lands heads in a single toss.

iii)

P(AUB) refers to the probability that either event A  or event B occurs. This is read out as the probability of A union B. This is usually given by the sum of the individual probabilities.

iv)

1 - P(ANB) is the probability that both events A  and B do not occur                                 together, but either  may occur by itself. P(ANB) is the probability that both events A  and B occur together. This is read out as the probability of A intersection B. Therefore implying that 1 - P(ANB) is simply the probability that either event A or B occurs but A and B can not occur together.

v)

1- P(AUB) refers to the probability that neither event A  or event B occurs. Earlier we defined P(AUB) as the probability that either event A  or event B occurs. 1- P(AUB) simply the complement of P(AUB).

vi)

P(AIB) refers to the probability that event A occurs  given the fact that event B occurs. This is a conditional probability event which evaluates the likelihood of an event A occurring given that an associated event B has already occurred

f(1986) = 0

Step-by-step explanation:

f(a + b) = f(a) + f(b) - 2f(ab)

We need to find f(1986)

f(1986) = f(1 +1985). Using the above formula, we can write:

f(1 +1985) = f(1) + f(1985) - 2f(1 x 1985)

f(1986) = 1 - f(1985)                                         Equation 1

Applying the same formula again on f(1985), we get:

f(1985) = f(1 + 1984) = f(1) + f(1984) - 2f(1984)

f(1985) = 1 - f(1984)

Using this value in Equation 1, we get:

f(1986) = 1 - (1 - f(1984))

f(1986)= f(1984)

Continuing this, we can observe,

f(1986) = f(1984) = f(1982) = f(1980) = f(4) = f(2)

So,

f(1986) = f(2)

f(2) = f(1 + 1) = f(1) +f(1) - 2(1 x 1) = f(1) + f(1) - 2f(1)

f(2) = 1 + 1 - 2 = 0

Therefore,

f(1986) = f(2) = 0

f(1986) = 0

Step-by-step explanation:

f(a + b) = f(a) + f(b) - 2f(ab)

We need to find f(1986)

f(1986) = f(1 +1985). Using the above formula, we can write:

f(1 +1985) = f(1) + f(1985) - 2f(1 x 1985)

f(1986) = 1 - f(1985)                                         Equation 1

Applying the same formula again on f(1985), we get:

f(1985) = f(1 + 1984) = f(1) + f(1984) - 2f(1984)

f(1985) = 1 - f(1984)

Using this value in Equation 1, we get:

f(1986) = 1 - (1 - f(1984))

f(1986)= f(1984)

Continuing this, we can observe,

f(1986) = f(1984) = f(1982) = f(1980) = f(4) = f(2)

So,

f(1986) = f(2)

f(2) = f(1 + 1) = f(1) +f(1) - 2(1 x 1) = f(1) + f(1) - 2f(1)

f(2) = 1 + 1 - 2 = 0

Therefore,

f(1986) = f(2) = 0

Exercise 1. When answering a question with a full sentence, both in English, as in Spanish, you must use the context of the sentence and answer with the same ideas, in a logical way. Understand the context clues in the question, and the main idea, to answer the question, and use the appropriate vocabulary, as needed.

1. ¿Dónde están Pablo y Julio? Pablo y Julio están jugando en la parte de atrás de la casa

2. ¿Qué tienen los muchachos en las manos?; Los muchachos tienen los regalos de navidad en las manos.

3. En qué ciudad vive Carmen Duarte Guzmán?; Carmen Duarte Guzmán vive en Ciudad de México.

4. ¿Qué recibió Julio de su amiga Iris-Teresa?; Julio recibió un regalo de su amiga Iris-Teresa.

5. ¿Qué le dijo Iris-Teresa?; Iris-Teresa le dijo que no los podía abrir aún.

6. ¿Dónde hay computadoras?; Hay computadoras en las empresas.

7. ¿Dónde va Pablo esta tarde?; Pablo va a su casa a descansar.

8. ¿Por qué va Julio a telefonear a José Antonio?; Julio va a telefonear a José Antonio porque está de cumpleaños.

9. ¿Por qué no llama nunca Pablo cuando está de vacaciones?; Pablo no llamada cuando está de vacaciones porque se quiere desconectar.

10. ¿Por qué no le cuesta mucho a Julio hacer llamadas?; No le cuesta mucho a Julio hacer llamadas porque tiene un buen plan de telefonía.

Exercise 2. Use of superlatives. Same as in English. It would be the most, in a comparative scenario. Thus:

1. Guapo: Guapísimo. 2. Son altas: son altísimas. 3. Es fea: Es feísima. 4. Están contentos: Están contentísimos. 5. Es rápido: es rapidísimo.