Question

Question:

include the MATLAB output and commands used with each problem

Generate two random 10 × 10 matrices with numbers between -10 and 10. This can be done with

>> A = randi([-10,10],10,10)

>> B = randi([-10,10],10,10)

1.     with MATLAB to determi whether A and B are invertible matrices .

2.     If A is invertible, use MATLAB to show that A^-1A = I

3.     Determi whether (AB)^-1 = B^-1A^-1

4.     Determin whether (A^T)^-1 = (A^-1)^T

5.     Determe whether (A³)^-1 = (A^-1)³

6.     Determi whether (A + B)(A - B) = A² - B²

Answer 1

Solution:- given question is

Explaination:-(2)

The inverse of a matrix A is denoted by A−1such that the following relationship holds −

AA−1 = A−1A = 1

The inverse of a matrix does not always exist. If the determinant of the matrix is zero, then the inverse does not exist and the matrix is singular.

Inverse of a matrix in MATLAB is calculated using the inv function. Inverse of a matrix A is given by inv(A).

Example

Create a script file and type the following code −

Live Demo

a = [ 1 2 3; 2 3 4; 1 2 5]
inv(a)

When you run the file, it displays the following result −

a =
   1     2     3
   2     3     4
   1     2     5
ans =
   -3.5000    2.0000    0.5000
   3.0000   -1.0000    -1.0000
   -0.5000      0       0.5000

Explaination:- (3)

such matrices exist. Note first that for invertible A,BA,B we have (AB)−1=A−1B−1(AB)−1=A−1B−1 if and only if AB=BAAB=BA. Thus, this comes down to finding a collection of invertible matrices which commute.

The simplest non-trivial set of such matrices is the set of diagonal matrices with all non-zero diagonal entries.

A group GG in which (ab)−1=a−1b−1(ab)−1=a−1b−1for all a,b∈Ga,b∈G is abelian. There's plenty of groups of invertible matrices that are abelian.

A very simple example is the group of powers of AA, where AA is an invertible matrix. Another important one is the set of matrices of the form

[a−bba][ab−ba]

with aa and bb real, a2+b2≠0a2+b2≠0.

Explaination:-(6)

Let us calculate (A−B)(A+B)(A−B)(A+B) as follows using the fact that the matrix product is distributive.

(A−B)(A+B)=A(A+B)−B(A+B)=A2+AB−BA−B2=A2−B2+(AB−BA).(A−B)(A+B)=A(A+B)−B(A+B)=A2+AB−BA−B2=A2−B2+(AB−BA).

Thus if (A−B)(A+B)=A2−B2(A−B)(A+B)=A2−B2 then AB−BA=OAB−BA=O, the zero matrix. Equivalently, AB=BAAB=BA.

Note that matrix multiplication is not commutative, namely, AB≠BAAB≠BA in general.
Thus we can disprove the statement if