Let A be a non singular matrix of order n × n, Then | adj (3A)| is equal to:

$$A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$A=[cosθsinθ​−sinθcosθ​]

For a matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$[ac​bd​], the inverse is:

$$\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$ad−bc1​[d−c​−ba​]

The determinant of $A$A:

$$\text{det}(A) = \cos\theta \cdot \cos\theta - (-\sin\theta) \cdot \sin\theta$$det(A)=cosθ⋅cosθ−(−sinθ)⋅sinθ

$$= \cos^2\theta + \sin^2\theta$$=cos2θ+sin2θ

$$= 1$$=1

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The inverse of $A$A:

$$A^{-1} = \frac{1}{1}\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$A−1=11​[cosθ−sinθ​sinθcosθ​]

$$A^{-1} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$A−1=[cosθ−sinθ​sinθcosθ​]

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The transpose of $A^{-1}$A−1 flips the matrix along its main diagonal:

$$(A^{-1})^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$(A−1)T=[cosθsinθ​−sinθcosθ​]

This equals the original matrix $A$A.

Therefore, $(A^{-1})^T = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$(A−1)T=[cosθsinθ​−sinθcosθ​]