GATE Chemical 2018 | Question: 47

Question

For the matrix $A=\begin{bmatrix} \cos\theta &-\sin\theta \\ \sin \theta & \cos\theta \end{bmatrix}$ if $\det$ stands for the determinant and $A^{T}$ is the transpose of $A$ then the value of $\det(A^{T}\:A)$ is _____________

Answer 1

 

The problem asks for the value of $\det(A^T A)$. Based on the properties of determinants:

• $\det(M N) = \det(M) \det(N)$

• $\det(M^T) = \det(M)$

Therefore:

$$\det(A^T A) = \det(A^T) \det(A) = \det(A) \det(A) = (\det(A))^2$$

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Step 1: Calculate the Determinant of A

The given matrix is a standard rotation matrix:

$$A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$

Calculate the determinant using the $2 \times 2$ formula ($ad - bc$):

$$\det(A) = (\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta)$$

$$\det(A) = \cos^2 \theta + \sin^2 \theta$$

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Step 2: Apply Trigonometric Identity

From fundamental trigonometry, we know that for any angle $\theta$:

$$\cos^2 \theta + \sin^2 \theta = 1$$

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Step 3: Find the Final Value

Substitute the value of $\det(A)$ back into our equation from Step 1:

$$\det(A^T A) = (1)^2 = 1$$

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Final Answer:

1