CSIR JUNE 2011 PART B QUESTION 32 SOLUTION (Derivative of the determinat map evaluated at $(H,K)$)

For $V = (V_1,V_2) \in \Bbb R^2$ and $W = (W_1,W_2)$, consider the determinant map $det:\Bbb R^2 \times \Bbb R^2 \to \Bbb R$ defined by $det(V,W) = V_1W_2 - V_2W_1$. Then the derivative of the determinant map at $(V,W) \in \Bbb R^2 \times \Bbb R^2$ evaluated on $(H,K) \in \Bbb R^2 \times \Bbb R^2$ is
1) $det(H,W) + det(V,W)$,
2) $det(H,K)$,
3)$det(H,V) + det(W,K)$,
4) $det(V,H) + det(K,W)$.
Solution: We want to find the derivative of the determinant map at the point $(H,K)$, that is along the vector $(H,K)$. So we need to find the directional derivative of determinant along the point $(H,K)$. 

Now, the directional derivative of the determinant map along $(H, K)$ is given by $\nabla(det) \cdot (H,K)$ where $\cdot$ is the dot product. 
First, we shall calculate the gradient $\nabla(det)$:
$$\big(\frac{\partial(det)}{\partial V_1},\frac{\partial(det)}{\partial V_2},\frac{\partial(det)}{\partial W_1},\frac{\partial(det)}{W_2}\big).$$
and this is equal to $$(W_2,-W_1,-V_2,V_1)$$ 
Let $H=(H_1,H_2)$ and $K=(K_1,K_2)$. Now, to calculate the required directional derivative, we have to take the inner product of the above gradient with the vector $(H_1, H_2, K_1,K_2)$ and this gives $W_2H_1-W_1H_2-V_2K_1+V_1K_2 = (H_1W_2-H_2W_1)+ (V_1K_2-V_2K_1)$ and this is equal to $$det(H,W) + det(V,K).$$ So option (1) is correct.

Share to your groups:
FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS AND COMMENTS BELOW IN THE COMMENTS SECTION. ALSO, SUGGEST PROBLEMS TO SOLVE.

No comments:

Post a Comment

Featured Post

NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$

Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...

Popular Posts