if a is an orthogonal matrix then show that A^t and A^-1 are orthogonal

 An orthogonal matrix is a square matrix whose transpose is equal to its inverse. In other words, if matrix A is orthogonal, then A^T = A^(-1).

Let's prove this:

1. Start with the definition of an orthogonal matrix: A * A^T = I, where I is the identity matrix.

2. Take the transpose of both sides: (A * A^T)^T = I^T.

3. Using the property of matrix transpose (transpose of a product is the product of transposes in reverse order): A^T * (A^T)^T = I.

4. Since (A^T)^T = A, we have A^T * A = I.

5. Now, multiply both sides of the equation by A^(-1): A^T * A * A^(-1) = I * A^(-1).

6. Since A^T * A = I, the left-hand side becomes A^(-1) = I * A^(-1).

7. Thus, A^T = A^(-1), which means if A is an orthogonal matrix, then its transpose and its inverse are the same, making both A^T and A^(-1) orthogonal matrices.

This completes the proof.

1.Show that the matrix A is periodic or not

A=    | 1     -2    -6|

         | -3     2     9|

         |2     0     - 3|

To determine if matrix A is periodic, we need to check if there exists a positive integer 'k' such that A^k = A, where '^' denotes matrix exponentiation. In other words, if raising the matrix A to the power of 'k' results in the same matrix A, then A is periodic with period 'k'.

Calculate the powers of matrix A:

A^2 =  | 19    -8   -39|

         | -33   14   63 |

         | -12   4    21 |

A^3 =  | -121   52    237|

         | 195   -82  -375|

         | 54      -24  -108|

A^4 =  | 487    -208  -951|

         | -789   334   1533|

         | -216   92     423|

Since the calculated powers of A do not match the original matrix A, A is not periodic.

2.Let f(t)=t^2-3t-4 then prove that f(A)=0

Certainly! Let's proceed to calculate \(f(A)\) for the given matrix \(A\):

Matrix A:

```

| 1   2|

| 3   2|

```

The polynomial \(f(t) = t^2 - 3t - 4\).

To calculate \(f(A)\), we substitute matrix \(A\) into the polynomial \(f(t)\):

f(A)=A2−3A−4I

I=[10​01​

A2=A⋅A=[13​22​]⋅[13​22​]=[79​610​]

3A=3⋅[13​22​]=[39​66​]

f(A)=[79​610​]−[39​66​]−[40​04​]=[00​00​]

Since $�(�)$ is the zero matrix, we've shown that $�(�)=0$