Question

If \(A = \frac{1}{2}[\begin{array}{cc}-1 & -\sqrt{3} \\ \sqrt{3} & -1\end{array}]\) then \(A^{-1}-A^2\) is not

• A. a null-matrix
• B. a unit matrix
• C. a diagonal matrix
• D. a scalar matrix

Hint:

A is a rotation matrix by $120^{\circ}$.

Answer 1

The Correct Option is B

Step 1: Recognise \(A\)
\(A=\begin{bmatrix}-\frac12&-\frac{\sqrt3}{2}\\\frac{\sqrt3}{2}&-\frac12\end{bmatrix}=\begin{bmatrix}\cos120^{\circ}&-\sin120^{\circ}\\\sin120^{\circ}&\cos120^{\circ}\end{bmatrix}\), a rotation by \(120^{\circ}\).

Step 2: Find \(A^2\) and \(A^{-1}\)
\(A^2\) is a rotation by \(240^{\circ}\). \(A^{-1}\) is a rotation by \(-120^{\circ}\), which is the same as \(240^{\circ}\). So \(A^{-1}=A^2\).

Step 3: Conclude
\(A^{-1}-A^2=O\), the null matrix. A null matrix is also diagonal and scalar (with scalar \(0\)), but it is not a unit matrix because its diagonal entries are not \(1\). So the answer is (B).

Final Answer:
The difference is the null matrix, so it is not a unit matrix, option (B). \[ \boxed{\text{(B)}} \]