Question

If \( A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \), \( P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) and \( X = A P A^T \), then \( A^T X^{50} A = \) 
 

• A. \(\begin{pmatrix} 0 & 1 1 & 0 \end{pmatrix}\)
• B. \(\begin{pmatrix} 2 & 1 0 & -1 \end{pmatrix}\)
• C. \(\begin{pmatrix} 25 & 1 1 & -25 \end{pmatrix}\)
• D. \(\begin{pmatrix} 1 & 50 0 & 1 \end{pmatrix}\)

Hint:

For any matrix of the form \(\begin{pmatrix} 1 & a 0 & 1 \end{pmatrix}\), the \(n^{th}\) power is simply \(\begin{pmatrix} 1 & na 0 & 1 \end{pmatrix}\). This is a very common shortcut in matrix algebra!

Answer 1

The Correct Option is D

To solve the problem, we need to compute \(A^T X^{50} A\) where \(A\), \(P\), and \(X\) are given as follows:

• \( A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)
• \( P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)
• \( X = A P A^T \)

Let's start by finding \(X\):

\[ A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad A^T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \]

Now, calculate \(PA^T\):

\[ PA^T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \]

Next, calculate \(AP\):

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

Now compute \(X = APA^T\):

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

Next, we find \(X^{50}\). Notice that:

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

Since \(X^n = \begin{pmatrix} 1 & -n \\ 0 & 1 \end{pmatrix}\) due to the exponent pattern of the upper triangular matrix:

\[ X^{50} = \begin{pmatrix} 1 & -50 \\ 0 & 1 \end{pmatrix} \]

Finally, calculate \(A^T X^{50} A\):

\( A^T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)

\[ A^T X^{50} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & -50 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -50 \\ 0 & -1 \end{pmatrix} \]

Then calculate \(A^T X^{50} A\):

\[ \begin{pmatrix} 1 & -50 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 50 \\ 0 & 1 \end{pmatrix} \]

Thus, the solution to the problem is:
\[\begin{pmatrix} 1 & 50 \\ 0 & 1 \end{pmatrix}\]

This matches the fourth option in the given options.