Question

The components of pure shear strain in a sheared material are given in the matrix form:
\[ \epsilon = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \] Here, \(\text{Trace}(\epsilon) = 0\). Given, P = \text{Trace}(\epsilon^8) and Q = \text{Trace}(\epsilon^{11}). \text{The numerical value of} (P + Q) \text{ is} _________ \text{. (in integer)}.

Hint:

For powers of matrices, use properties such as \( \text{Trace}(A^n) = \text{sum of eigenvalues of } A^n \).

Answer 1

Correct Answer: 32

First, we calculate the matrix powers of \( \epsilon \):
\[ \epsilon^2 = \begin{bmatrix} 1 & 1
1 & -1 \end{bmatrix} \times \begin{bmatrix} 1 & 1
1 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0
0 & 2 \end{bmatrix} \] \[ \epsilon^3 = \epsilon \times \epsilon^2 = \begin{bmatrix} 1 & 1
1 & -1 \end{bmatrix} \times \begin{bmatrix} 2 & 0
0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 2
2 & -2 \end{bmatrix} \] \[ \epsilon^4 = \epsilon^2 \times \epsilon^2 = \begin{bmatrix} 2 & 0
0 & 2 \end{bmatrix} \times \begin{bmatrix} 2 & 0
0 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 0
0 & 4 \end{bmatrix} \] Since \( \epsilon^8 = \epsilon^4 \times \epsilon^4 \), we get:
\[ \epsilon^8 = \begin{bmatrix} 16 & 0
0 & 16 \end{bmatrix} \] Thus, the trace of \( \epsilon^8 \) is:
\[ \text{Trace}(\epsilon^8) = 16 + 16 = 32 \] Now, for \( \epsilon^{11} = \epsilon^8 \times \epsilon^2 \times \epsilon \):
\[ \epsilon^{11} = \begin{bmatrix} 16 & 0
0 & 16 \end{bmatrix} \times \begin{bmatrix} 2 & 0
0 & 2 \end{bmatrix} \times \begin{bmatrix} 1 & 1
1 & -1 \end{bmatrix} \] Thus, we find:
\[ P = 32, \quad Q = 0 \] Therefore, the value of \( P + Q \) is:
\[ P + Q = 32 + 0 = 32 \]