If X=`[{:(3,-4),(1,-1):}]`, then value of `X^(n)` is (where n is a natural number)

Options

A. `{:[(,3n,-4),(,n,-n)]:}`
B. `{:[(,2n+n,5-n),(,n,-n)]:}`
C. `{:[(,3^(n),(-4)^(n)),(,1^(n),(-1)^(n))]:}`
D. `{:[(,2n+1,-4),(,n,-(2n-1))]:}`

Answer

To find the value of \( X^n \) where \( X = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \) and \( n \) is a natural number, we can use the property of matrices and their eigenvalues.

### Step 1: Find the characteristic polynomial of the matrix \( X \).

The characteristic polynomial is given by the determinant of \( (X - \lambda I) \), where \( I \) is the identity matrix.

\[ X - \lambda I = \begin{pmatrix} 3 - \lambda & -4 \\ 1 & -1 - \lambda \end{pmatrix} \]

Now, we calculate the determinant:

\[ \text{det}(X - \lambda I) = (3 - \lambda)(-1 - \lambda) - (-4)(1) \]

Calculating this gives:

\[ = (-3 - 3\lambda + \lambda + \lambda^2 + 4) = \lambda^2 - 2\lambda + 1 \]

### Step 2: Solve the characteristic polynomial.

Setting the characteristic polynomial to zero:

\[ \lambda^2 - 2\lambda + 1 = 0 \]

This can be factored as:

\[ (\lambda - 1)^2 = 0 \]

Thus, the eigenvalue \( \lambda = 1 \) with algebraic multiplicity 2.

### Step 3: Find the eigenvectors.

To find the eigenvectors, we solve \( (X - I)v = 0 \):

\[ X - I = \begin{pmatrix} 3 - 1 & -4 \\ 1 & -1 - 1 \end{pmatrix} = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix} \]

Setting up the equations:

\[ 2x - 4y = 0 \quad \text{(1)} \]
\[ x - 2y = 0 \quad \text{(2)} \]

From equation (2), we can see that \( x = 2y \). Let \( y = t \), then \( x = 2t \). The eigenvector corresponding to \( \lambda = 1 \) is:

\[ v = \begin{pmatrix} 2t \\ t \end{pmatrix} = t \begin{pmatrix} 2 \\ 1 \end{pmatrix} \]

### Step 4: Use the Jordan form for powers of \( X \).

Since \( X \) has a repeated eigenvalue, we can express \( X \) in Jordan form. The Jordan form \( J \) is:

\[ J = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]

The matrix \( X \) can be expressed as \( X = PJP^{-1} \), where \( P \) is the matrix of eigenvectors.

### Step 5: Calculate \( X^n \).

Using the property of the Jordan block, we have:

\[ J^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \]

Thus,

\[ X^n = PJ^nP^{-1} \]

### Step 6: Final Calculation.

After calculating \( X^n \), we find:

\[ X^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \]

Substituting back gives us the final result for \( X^n \).

### Conclusion

The value of \( X^n \) can be expressed in terms of \( n \) as follows:

\[ X^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \]