Question

If \(A = \begin{bmatrix} 0 & -\tan(\alpha/2) \\ \tan(\alpha/2) & 0 \end{bmatrix}\) and \(I\) is the identity matrix of order 2, prove that \(I+A = (I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\).

Hint:

When dealing with trigonometric identities in matrices, using the half-angle tangent substitutions (\(t = \tan(\alpha/2)\)) is a very powerful technique that converts trigonometric expressions into rational algebraic expressions, which are often easier to handle in matrix multiplication.

Answer 1

Step 1: Understanding the Concept:
This problem involves proving a matrix identity. The strategy is to calculate the Left-Hand Side (LHS) and Right-Hand Side (RHS) separately and show that they are equal. The proof uses matrix addition, subtraction, multiplication, and the half-angle formulas for sine and cosine in terms of tangent.
Step 2: Key Formula or Approach:
1. Calculate \( \text{LHS} = I + A \).
2. Calculate \( I - A \).
3. Use the half-angle formulas: \( \cos\alpha = \frac{1-t^2}{1+t^2} \) and \( \sin\alpha = \frac{2t}{1+t^2} \), where \( t = \tan(\alpha/2) \).
4. Compute \( \text{RHS} = (I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \).
5. Compare LHS and RHS.
Step 3: Detailed Explanation:
Let \( t = \tan(\alpha/2) \). Then \[ A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Calculate LHS: \[ \text{LHS} = I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \] Calculate RHS: \[ I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \] Using the half-angle formulas: \[ \cos\alpha = \frac{1-t^2}{1+t^2}, \quad \sin\alpha = \frac{2t}{1+t^2} \implies \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t \\ 2t & 1-t^2 \end{bmatrix} \] Now multiply: \[ \text{RHS} = (I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t \\ 2t & 1-t^2 \end{bmatrix} \] \[ = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 + 2t^2 & -2t - t + t^3 \\ -t + t^3 + 2t & 2t^2 + 1-t^2 \end{bmatrix} = \frac{1}{1+t^2} \begin{bmatrix} 1+t^2 & -t(1+t^2) \\ t(1+t^2) & 1+t^2 \end{bmatrix} \] Divide each element by \(1+t^2\): \[ \text{RHS} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \] Compare LHS and RHS: \[ \text{LHS} = \text{RHS} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \] Step 4: Final Answer:
Since LHS = RHS, the identity is proven. ✅