Area after matrix transformation: A unit square is mapped to a parallelogram....

[ShanQ]

Here is the question.
The unit square is mapped to a parallelogram of area 3 by the matrix below
$$\textbf{B}=\begin{bmatrix} m&2\\m&m\end{bmatrix}$$Find the possible value of m.

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I have tried the question by myself as below.
My solution:
Set the original unit square as $$\begin{bmatrix} 0&1&0&1\\ 0&0&1&1 \end{bmatrix}$$The final image after transformation: $$\begin{bmatrix} m&2\\m&m\end{bmatrix} \begin{bmatrix} 0&1&0&1\\ 0&0&1&1 \end{bmatrix}= \begin{bmatrix} 0&m&2&m+2\\ 0&m&m&2m \end{bmatrix}$$
Given it is a parallelogram after transformation and its area = 3
$$m\times m=3$$m must be positive. $$m >0$$Therefore, we can obtain $$m=\sqrt3$$

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Here is the textbook solution below:
determinant of the transformation matrix: $$det=m^2-2m$$Area of image = determinant of the transformation matrix * the area of original region.
$$3= |m^2-2m|\times1$$We can obtain:
$$m^2-2m=\pm3$$
Case 1:
$$m^2-2m=3$$We can obtain the final answer: $$m_1=3, m_2=-1$$
Case 2:
$$m^2-2m=-3$$Since $$\Delta =b^2-4ac=4-4(3)=-8<0$$There is no Real solution.

Conclusion:
$$m_1=3, m_2=-1$$

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If m can be negative, I would modify my answer $$m=\pm \sqrt3$$Could you please help with this question? What mistake have I made?
Thank you.

 

[blamocur]

Why did you write $m \times m = 3$ ?

 

[Harry_the_cat]

The textbook solution is correct (You can't always count on that!) and IMO the most efficient way to do the problem.

Doing it your way, if you plot the points to form the parallelogram image, the area is NOT $\displaystyle m\times m$. (I assume that is what you are thinking when you write $\displaystyle m\times m =3$.

The horizontal distance between $\displaystyle (2, m)$ and $\displaystyle (m, m)$ is $\displaystyle |m-2|$. The vertical height is $\displaystyle |m|$.

This means the area is $\displaystyle |m(m-2)|$.

(Note that $\displaystyle m(m-2)$ is equal to the determinant.]

 

[ShanQ]

Thank you so much for your help. Fully understand now. Cheers.