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Part c) I'm not quite sure what to do, I've found the det(U) is 2, but no idea what this actually shows to be honest, any help?
Matrices & Geometric Transformations
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Part c) I'm not quite sure what to do, I've found the det(U) is 2, but no idea what this actually shows to be honest, any help?
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The most direct thing to do is apply the two matrices to each of the vectors (0, 0), (1, 0), (0, 1), and (1, 1) to determine the new figure. Then find the area of that. There are also theorems relating the determinants of the transformation matrices to the area.
synkk
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HallsofIvy said:
Thank you, I've done that and got the right answer however in the solutions it simply states determinant of RS = 2, therefore area scale factor of U is 2, therefore image of U has area 2. I don't understand what the determinant has to do with it, could you explain please?
I mean det(RS), my apologies, thanks.vela said:
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Then clearly the author of your text expects you to know that the area of a two dimensional object, when it transformed by A, is multiplied by det(A).
If [tex]A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] is applied to the vector (1, 0), we have, of course, (a, c) and if it is applied to the vector (0, 1), we have (b, d). That is, the square having (1, 0) and (0, 1) as sides (and so area 1), is transformed into the parallelogram having (a, c) and (b, d) as sides. And, of course, the area of that parallelogram is the length of the cross product of those two vectors (taking [itex]\vec{k}[/itex] to be 0 to make it three dimensional). That cross product is
[tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a & c & 0 \\ b & d & 0 \end{array}\right|= (ad- bc)\vec{k}[/tex] and has length ad- bc, the determinant of A.
If [tex]A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex] is applied to the vector (1, 0), we have, of course, (a, c) and if it is applied to the vector (0, 1), we have (b, d). That is, the square having (1, 0) and (0, 1) as sides (and so area 1), is transformed into the parallelogram having (a, c) and (b, d) as sides. And, of course, the area of that parallelogram is the length of the cross product of those two vectors (taking [itex]\vec{k}[/itex] to be 0 to make it three dimensional). That cross product is
[tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a & c & 0 \\ b & d & 0 \end{array}\right|= (ad- bc)\vec{k}[/tex] and has length ad- bc, the determinant of A.