Here is the question.
The unit square is mapped to a parallelogram of area 3 by the matrix below
[math]\textbf{B}=\begin{bmatrix} m&2\\m&m\end{bmatrix}[/math]Find the possible value of m.
I have tried the question by myself as below.
My solution:
Set the original unit square as [math]\begin{bmatrix} 0&1&0&1\\ 0&0&1&1 \end{bmatrix}[/math]The final image after transformation: [math]\begin{bmatrix} m&2\\m&m\end{bmatrix} \begin{bmatrix} 0&1&0&1\\ 0&0&1&1 \end{bmatrix}= \begin{bmatrix} 0&m&2&m+2\\ 0&m&m&2m \end{bmatrix}[/math]
Given it is a parallelogram after transformation and its area = 3
[math]m\times m=3[/math]m must be positive. [math]m >0[/math]Therefore, we can obtain [math]m=\sqrt3[/math]
Here is the textbook solution below:
determinant of the transformation matrix: [math]det=m^2-2m[/math]Area of image = determinant of the transformation matrix * the area of original region.
[math]3= |m^2-2m|\times1[/math]We can obtain:
[math]m^2-2m=\pm3[/math]
Case 1:
[math]m^2-2m=3[/math]We can obtain the final answer: [math]m_1=3, m_2=-1[/math]
Case 2:
[math]m^2-2m=-3[/math]Since [math]\Delta =b^2-4ac=4-4(3)=-8<0[/math]There is no Real solution.
Conclusion:
[math]m_1=3, m_2=-1[/math]
If m can be negative, I would modify my answer [math]m=\pm \sqrt3[/math]Could you please help with this question? What mistake have I made?
Thank you.
The unit square is mapped to a parallelogram of area 3 by the matrix below
[math]\textbf{B}=\begin{bmatrix} m&2\\m&m\end{bmatrix}[/math]Find the possible value of m.
I have tried the question by myself as below.
My solution:
Set the original unit square as [math]\begin{bmatrix} 0&1&0&1\\ 0&0&1&1 \end{bmatrix}[/math]The final image after transformation: [math]\begin{bmatrix} m&2\\m&m\end{bmatrix} \begin{bmatrix} 0&1&0&1\\ 0&0&1&1 \end{bmatrix}= \begin{bmatrix} 0&m&2&m+2\\ 0&m&m&2m \end{bmatrix}[/math]
Given it is a parallelogram after transformation and its area = 3
[math]m\times m=3[/math]m must be positive. [math]m >0[/math]Therefore, we can obtain [math]m=\sqrt3[/math]
Here is the textbook solution below:
determinant of the transformation matrix: [math]det=m^2-2m[/math]Area of image = determinant of the transformation matrix * the area of original region.
[math]3= |m^2-2m|\times1[/math]We can obtain:
[math]m^2-2m=\pm3[/math]
Case 1:
[math]m^2-2m=3[/math]We can obtain the final answer: [math]m_1=3, m_2=-1[/math]
Case 2:
[math]m^2-2m=-3[/math]Since [math]\Delta =b^2-4ac=4-4(3)=-8<0[/math]There is no Real solution.
Conclusion:
[math]m_1=3, m_2=-1[/math]
If m can be negative, I would modify my answer [math]m=\pm \sqrt3[/math]Could you please help with this question? What mistake have I made?
Thank you.
