MHTCET 11 Apr 2026 Shift 1 Paper

Concept:$A$ is a rotation matrix by $120^\circ$, so $A^3 = I$.Explanation:We have $A = \frac{1}{2}\begin{pmatrix} -1 & -\sqrt{3} \\ \sqrt{3} & -1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}$.This matches the rotation matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$.Hence $\cos\theta = -\frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$, so $\theta = 120^\circ$.Thus applying $A$ three times rotates by $3 \times 120^\circ = 360^\circ$, giving $A^3 = I$.Multiplying $A^3 = I$ on the left by $A^{-1}$ yields $A^2 = A^{-1}$.Therefore $A^{-1} - A^2 = A^{-1} - A^{-1} = 0$.The zero matrix $0$ is a null matrix, a diagonal matrix, and also a scalar matrix because $0 = 0 \cdot I$.But it is not the unit matrix $I$, since $I$ has diagonal entries $1$, not $0$.Answer:Option B: a unit matrix.