In this exercise we will show that the absolute value of the determinant of a 2

Question

In this exercise we will show that the absolute value of the determinant of a 2 × 2 matrix gives the area of the parallelogram determined by its columns. (a) Let , e R. Prove the Lagrange identity: llu x 2-1 12111F-(7.7)2 (b) Using the Lagrange identity, show that IF ×1-/71111ll sin , where is the angle between Tt and (c) Using part (b) explain why the area of the parallelogram defined by u and has area llTx Include a diagram in your explanation. (d) Let u-l bl and = | d | . Compute x and show 11 × 11-1ad-bel. (e) Let T : R2 R2 be a linear transformation and let T( ) A , where A is a 2 x 2 matrix. iD 0 Explain why Idet (A)| is the area of the parallelogram defined by T( ) and T(72)

Explanation / Answer

Ans(a):
To prove:
||u x v||^2=||u||^2||v||^2-(u*v)^2

Start from the right side:
||u||^2||v||^2-(u*v)^2
=||u||^2||v||^2-(||u|| ||v^2||cos(theta))^2 {using definition of dot product}
=||u||^2||v||^2-||u||^2||v^2||cos^2(theta)
=||u||^2||v||^2(1-cos^2(theta))
=||u||^2||v||^2 sin^2(theta)
=(||u||||v|| sin(theta))^2 {using definition of cross product}
=||u x v||^2

Hence proved.

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