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

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

Step 1: Find the inverse
$\det A=\frac14(1+3)=1$. For a $2\times2$ matrix with determinant $1$, the inverse swaps the diagonal entries and negates the off-diagonal ones.
So $A^{-1}=\frac12\begin{bmatrix}-1&\sqrt3\\-\sqrt3&-1\end{bmatrix}$.

Step 2: Square $A$
$A^2=\frac14\begin{bmatrix}1-3&2\sqrt3\\-2\sqrt3&-2\end{bmatrix}=\frac12\begin{bmatrix}-1&\sqrt3\\-\sqrt3&-1\end{bmatrix}$.
This equals $A^{-1}$, so $A^{-1}-A^2=O$. It is null, diagonal and scalar, but not a unit matrix. Option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]

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

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The Correct Option is B

Solution and Explanation

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