91Ó°ÊÓ

The transformation \(T: x=g(u,v), y=h(u, v)\) is given to be linear in \(u\) and \(v\). This means that the partial derivatives of \(g(u, v)\) and \(h(u, v)\) with respect to \(u\) and \(v\) will be constants. Let's compute the Jacobian matrix: $$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} $$ Since all partial derivatives are constant, the Jacobian will also be constant. Therefore, the statement is true.

The transformation is given as \(x=au+bv, y=cu+dv\). We need to determine if it maps triangular regions to triangular regions. Let's consider 3 arbitrary points in the \(u-v\) plane: \((u_1, v_1), (u_2, v_2), (u_3, v_3)\). When applying the transformation, we obtain the points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\): $$ x_1 = a u_1+b v_1, \: y_1 = c u_1+dv_1 \\ x_2 = a u_2+b v_2, \: y_2 = c u_2+dv_2 \\ x_3 = a u_3+b v_3, \: y_3 = c u_3+dv_3 $$ Since the transformation is linear, it will preserve the property that all three points are connected by straight lines. Hence, the resulting region will also be a triangle. Therefore, this statement is true.

The transformation is given as \(x=2v, y=-2u\). We need to determine if it maps circles to circles. Let's take the equation of a circle in the \(u-v\) plane, centered at \((u_0, v_0)\) with radius \(r\): $$ (u-u_0)^2+(v-v_0)^2=r^2 $$ Applying the transformation, we get: $$ x=2(v-v_0) \: \Rightarrow\: v=\frac{x}{2}+v_0 \\ y=-2(u-u_0) \: \Rightarrow\: u=-\frac{1}{2}y+u_0 $$ Writing the equation of the mapped curve in the terms of \(x\) and \(y\): $$ \left(-\frac{1}{2}y-u_0\right)^2+\left(\frac{x}{2}+v_0\right)^2=r^2 $$ Simplifying it we get: $$ \frac{1}{4}(y^2+x^2)+u_0y+v_0x = r^2 $$ This equation is not the equation of a circle in the \(x-y\) plane because it has a linear term in \(x\) and \(y\). Therefore, the statement is false. The transformation does not map circles to circles.