Question

If \(\mathrm{A}=\left(\begin{array}{cc}0 & 2009 \\ -2009 & 0\end{array}\right)\), then find \(\left[\mathrm{A}^{2009}+\left(\mathrm{A}^{\mathrm{T}}\right)^{2009}\right] .\) (1) \(\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) (2) \(\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)\) (3) \(\left(\begin{array}{cc}0 & 2009 \\ -2009 & 0\end{array}\right)\) (4) None of these

Short answer

Answer: The value of \([A^{2009} + (A^T)^{2009}]\) for the given matrix A is \(\left(\begin{array}{ll}0 & 0 \\\ 0 & 0\end{array}\right)\).

Step-by-step solution

Find the transpose of A

To find the transpose of A, we will simply swap the rows with the columns: \(A^T = \left(\begin{array}{cc}0 & -2009 \\\ 2009 & 0\end{array}\right)\).

Calculate A raised to the power of 2009 and (A^T) raised to the power of 2009

Since A is an anti-symmetric matrix \((A = -A^T)\), it follows that: \(A^2 = A * A = A * (-A^T) = 2009^2\left(\begin{array}{cc}1 & 0 \\\ 0 & 1\end{array}\right)\) So, \(A^{2009} = A * A^2 \cdot A^2 \cdot ... \cdot A^2\) = \(A * (A^2)^{1004}\) = \(A * (2009^2)^{1004}\left(\begin{array}{cc}1 & 0 \\\ 0 & 1\end{array}\right)\) = \(\left(\begin{array}{cc}0 & 2009 \\\ -2009 & 0\end{array}\right)\) Similarly, \((A^T)^{2009} = \left(\begin{array}{cc}0 & -2009 \\\ 2009 & 0\end{array}\right)\)

Add the two matrices

Now we add the two matrices obtained in the previous step: \([A^{2009} + (A^T)^{2009}] = \left(\begin{array}{cc}0 & 0 \\\ 0 & 0\end{array}\right)\)

Compare the result with the provided options

The result obtained from adding the two matrices matches with option (2): \(\left(\begin{array}{ll}0 & 0 \\\ 0 & 0\end{array}\right)\). Therefore, the correct answer is option (2).

Key concepts

Transpose of a Matrix

The transpose of a matrix is a fundamental operation in matrix algebra. To obtain the transpose of a matrix, you simply swap its rows with columns. For example, if you have a matrix \( A \) with elements \( a_{ij} \), the transpose of \( A \), denoted by \( A^T \), will have elements \( a_{ji} \). This means, if the original matrix is:

• \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \)

The transpose of this matrix, \( A^T \), would be:

• \( A^T = \begin{pmatrix} a & c \ b & d \end{pmatrix} \)

For the original problem's matrix:

• \( A = \begin{pmatrix} 0 & 2009 \ -2009 & 0 \end{pmatrix} \)

The transpose, \( A^T \), is:

• \( A^T = \begin{pmatrix} 0 & -2009 \ 2009 & 0 \end{pmatrix} \)

Notice how the off-diagonal elements swapped their positions and signs, which is typical in transposing. Understanding the transpose is crucial in more complex matrix operations.

Anti-symmetric Matrix

An anti-symmetric matrix, also known as a skew-symmetric matrix, has a peculiar property: it is its own negative transpose. Simply put, \( A = -A^T \). This implies that when you transpose the matrix and multiply by -1, you obtain the original matrix.
Here's a key characteristic:

• All diagonal elements in an anti-symmetric matrix are zero. This is because \( a_{ii} = -a_{ii} \), which is only possible if \( a_{ii} = 0 \).

Given the matrix
\( A = \begin{pmatrix} 0 & 2009 \ -2009 & 0 \end{pmatrix} \), observe:

• When transposed: \( A^T = \begin{pmatrix} 0 & -2009 \ 2009 & 0 \end{pmatrix} \)
• Negative of transpose: \(-A^T = \begin{pmatrix} 0 & 2009 \ -2009 & 0 \end{pmatrix} \), which is the original matrix \( A \)

This anti-symmetry is crucial in the problem because it gives rise to specific properties when raising such matrices to a power, as the negative sign cancels out over even powers.

Matrix Exponentiation

Matrix exponentiation is the process of multiplying a matrix by itself a specified number of times. This is similar to raising numbers to a power. In mathematical notation, \( A^n \) denotes matrix \( A \) multiplied by itself \( n \) times.
When dealing with anti-symmetric matrices, interesting patterns arise. For instance, an odd power of an anti-symmetric matrix results in the matrix itself, while even powers are often multiples of the identity matrix. This is due to the canceling effect of the anti-symmetry properties over repeated multiplications.

• In our example, since \( A^2 = 2009^2 \cdot I \) (Identity Matrix), where \( I \) is the identity matrix \( \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \),
• Thus, \( A^{2009} \) results in the combination of these powers as described.

This outcome is integral as it helps simplify complex expressions in matrix exponentiation, specifically applying when solving for higher powers effectively like in this problem.

Identity Matrix

The identity matrix is an essential concept in matrix algebra. It acts as a neutral element in matrix multiplication, much like the number 1 in regular arithmetic. For any matrix \( A \), when you multiply it with the identity matrix \( I \), the product is the matrix \( A \) itself:
\[ AI = IA = A \]The identity matrix for an \( n \times n \) matrix is a square matrix with ones on the diagonal and zeros elsewhere:

• For a \( 2 \times 2 \) matrix, \( I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \).

In the context of the problem, when calculating \( A^{2009} \), we found that even powers of the anti-symmetric matrix align with the structure of the identity matrix. The presence of identity matrices simplifies calculations, making them manageable by reducing terms effectively.
Every advanced matrix operation involves identity matrices in some way, emphasizing their indispensable nature in matrix mathematics.