Question

If $$ \mathrm{P}=\left|\begin{array}{cc} (\sqrt{3} / 2) & (1 / 2) \\ -(1 / 2) & (\sqrt{3} / 2) \end{array}\right|, \mathrm{A}=\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right| $$ and \(\mathrm{Q}=\mathrm{PAP}^{\mathrm{T}}\), Then \(\mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2013} \mathrm{P}=\ldots\) (c) \((1 / 4)\left|\begin{array}{cc}2+\sqrt{3} & 1 \\ -1 & 2-\sqrt{3}\end{array}\right|\) (d) (1/4) \(\left|\begin{array}{cc}2012 & 2+\sqrt{3} \\ 2+\sqrt{3} & 2012\end{array}\right|\)

Short answer

The short answer to the given question is: \(P^TQ^{2013}P=\frac{1}{4}\left|\begin{array}{cc}2012 & 2+\sqrt{3} \\\ 2+\sqrt{3} & 2012\end{array}\right|\) (which is option (d)).

Step-by-step solution

Find matrix Q using the given formula

We are given the formula for matrix Q: \(Q = PAP^T\). To find Q, we need to find the transpose of P, denoted as \(P^T\), and then perform the matrix multiplication. Given matrix P: $$ P=\left|\begin{array}{cc} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right| $$ To find P's transpose, switch the row and columns: $$ P^T = \left|\begin{array}{cc} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{array}\right| $$ Now, we must perform the matrix multiplication \(Q=PAP^T\) $$ Q = \left|\begin{array}{cc} \frac{3}{2} & \frac{1}{2} \\\ -\frac{1}{2} & \frac{3}{2} \end{array}\right| $$

Raise Q to the power of 2013

We need to raise Q to the power of 2013, which means \(Q^{2013} = Q * Q * ... * Q\) (2013 times). For any integer \(n \geq 1\), we observe that \(Q^n = PAP^T * PAP^T * ... * PAP^T\) (\(n\) times) \(= PA^nP^T\). Thus, we don't need to multiply \(Q\) to \(Q\) 2013 times, we just need to raise A to the power of 2013, then multiply by P and \(P^T\). Now, we need to find \(A^{2013}\). Since A is a simple matrix, we can easily observe that \(A^n = \left|\begin{array}{cc}1 & n \\0 & 1\end{array}\right|\). So, \(A^{2013} = \left|\begin{array}{cc}1 & 2013 \\0 & 1\end{array}\right|\).

Multiply the matrices to get the final answer

Now we need to perform the following matrix multiplication: \(P^TQ^{2013}P = P^T(PA^{2013}P^T)P\). We already found \(P^T\), \(P\), and \(A^{2013}\). First, we multiply \(A^{2013}\) by \(P^T\): $$ A^{2013} P^T = \left|\begin{array}{cc}1 & 2013 \\0 & 1\end{array}\right|\left|\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right|=\left|\begin{array}{cc}\frac{\sqrt{3}}{2}+1006.5 & \frac{1}{2}-1006.5 \\\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right| $$ Then, we multiply P with the result: $$ P^T(PA^{2013}P^T)P =\left|\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right|\left|\begin{array}{cc}\frac{1}{2}+\frac{\sqrt{3}}{2} & \frac{1}{2}-\frac{\sqrt{3}}{2} \\\ \frac{\sqrt{3}}{2}-\frac{1}{2} & -\frac{1}{2}-\frac{\sqrt{3}}{2}\end{array}\right|=\frac{1}{4}\left|\begin{array}{cc}2012 & 2+\sqrt{3} \\\ 2+\sqrt{3} & 2012\end{array}\right| $$ Thus, \(P^TQ^{2013}P=\frac{1}{4}\left|\begin{array}{cc}2012 & 2+\sqrt{3} \\\ 2+\sqrt{3} & 2012\end{array}\right|\) (which is option (d)).

Key concepts

Determinants and Matrices

In the realm of linear algebra, the determinant is an essential attribute of a square matrix. It is a scalar value that provides significant information about the matrix, such as whether the matrix is invertible, or its volume distortion factor during a linear transformation.

Determinants can be calculated for any square matrix, and when the determinant of a matrix is non-zero, it indicates that the matrix is invertible, which means there exists an inverse matrix for which matrix multiplication results in an identity matrix. Conversely, when the determinant is zero, it signifies that there is no inverse matrix.

For a 2x2 matrix \(A\), with elements \(a_{11}\), \(a_{12}\), \(a_{21}\), and \(a_{22}\), the determinant is calculated as \(a_{11}a_{22} - a_{12}a_{21}\). If we take the given matrix \(P\), the determinant can be found by multiplying the top left element by the bottom right element and subtracting the product of the top right and bottom left elements, resulting in \( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times -\frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1\), which assures us that matrix \(P\) is invertible.

When students engage with determinants and matrices, they not only learn about singular or non-singular matrices but also uncover important properties that aid in solving systems of linear equations and finding eigenvalues.

Matrix Transpose

The concept of transposition is another critical component within linear algebra. The transpose of a matrix \(A\), denoted as \(A^T\), is a new matrix whose rows are the columns of \(A\), and whose columns are the rows of \(A\). Essentially, what is being done is a switch between the rows and columns, maintaining the diagonal elements of the matrix intact.

For instance, if we have a 2x2 matrix like \(P\) given in the exercise, the transpose \(P^T\) is found by flipping the matrix over its main diagonal. The main diagonal is the one stretching from the top left to the bottom right corner. This results in the rows becoming columns and vice versa, giving us the transposed matrix \(P^T\) presented in the step by step solution.

Why Transpose a Matrix?

Transposing a matrix is useful in various mathematical computations and transformations, including solving systems of linear equations, orthogonalization of vectors, and in the context of matrix multiplications where the order of factors does not match up, like in the given problem where \(Q = PAP^T\). Understanding the transpose helps in finding solutions to such matrix equations.

Matrix Exponentiation

Applying exponentiation to matrices is slightly different than with regular numbers. Matrix exponentiation refers to multiplying a matrix by itself a certain number of times. For instance, \(A^2\) means \(AA\), and \(A^3\) means \(AAA\), and so on. It is crucial to remember that matrix multiplication is not commutative - the order in which the matrices are multiplied matters.

In the exercise, we need to compute \(Q^{2013}\). This does not involve multiplying the matrix \(Q\) 2013 times. Instead, properties of matrix multiplication allow us to simplify the process, especially when \(Q\) is a function of another matrix \(A\), as seen in \(Q = PAP^T\). Matrix exponentiation is potent as, under certain conditions, it can be simplified to the exponentiation of its eigenvalues, or like in the case of matrix \(A\) in our example, using its distinct structure to deduce a general formula for any power.

Application in Computation

Matrix exponentiation is commonly used in computer science for fast calculations, especially in algorithms involving graph theory and dynamic programming, where repeated matrix multiplication is necessary. The process exemplified in the solution demonstrates a way to compute high powers of matrices in a more efficient manner, where understanding the nature of the base matrix simplifies what initially seems like an overwhelming task.