[ MATLAB] Check some linear algebra rules: Enter the following matrices:
A =
[0 1
0 0]

B =

[1 2
-3 -6]

C =

[4 -2
-2 1]

The following rules of algebra hold for real numbers. However, most of them are false for matrices.
Use the matrices A;B;C above to perform the appropriate operations in MATLAB and determine
which rules are false. Below 0 denotes the 2 2 zero matrix. Note that if a rule holds for these
particular matrices, it does not mean that it is true in general. However, you should be able to
determine that from the theory you have learned in class.
(i) If BC = 0, then B = 0 or C = 0
(ii) If A2 = 0, then A = 0
(iii) (A + B)2 = A2 + 2AB + B2
(iv) (A - B)(A + B) = A^2 - B^2
(v) A(B + C) = AB + AC
(vi) A(B + C) = BA + CA
(vii) (AB)2 = A^2B^2

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:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

answer is : muchas/tipicas

have a nice day!

~recklesschicken