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Math Help - Determinants: What is the motivation?

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    Newbie fanofandrew's Avatar
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    Determinants: What is the motivation?

    The name 'determinant' implies that it determines. It determines the value of variable x3 by calculating its coefficient in the linear system of equations say, a1x1 + a2x2 + a3x3 = c1; a4x1 + a5x2 + a6x3 = c2; a7x1 + a8x2 + a9x3 = c3; using Gauss-elimination, where a1 through a9 are constants, x1, x2 and x3 are the variables whose values will give the solution to the system of equations, and c1, c2 and c3 are also constants though they form no part of the determinant. The determinant is represented as | (first row) a1 a2 a3 (second row) a4 a5 a6 (third row) a7 a8 a9 |. This is how I understand the motivation of determinants. In all textbooks however they give the standard formula without explaining the motivation, viz., that a1(a5.a9 - a8.a6) - a2(a4.a9 - a7.a6) + a3(a4.a8 - a7.a5) = 0 derives from cross-multiplication of the two-variate system a1x1 + a2x2 = a3; a4x1 + a5x2 = a6; a7x1 + a8x2 = a9. I don't see the similarity of steps or congruence between this and the Gauss-elimination method that I understand as the real motivation and that I stated first, though both methods give the same result. Could anyone illumine?
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    Re: Determinants: What is the motivation?

    could you ask your question without the wall of text?

    Are you asking what good determinants are?
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  3. #3
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    Re: Determinants: What is the motivation?

    No, just determinants in general. Why do we multiply, add and subtract the elements the way we do to get the value of the determinant? What is the motivation?
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    Re: Determinants: What is the motivation?

    The determinant of a matrix is the scale factor a shape/volume undergoes when the matrix transformation is applied onto that volume.

    I'll illustrate with the 2D case. Suppose we have two vectors, \vec{a} and \vec{b} that form the columns of the matrix \begin{bmatrix}a_1 & b_1 \\ a_2 & b_2\end{bmatrix}.

    Recall that the column vectors of the matrix are the images of the axis vectors \vec{i}=\begin{bmatrix}1 \\ 0 \end{bmatrix} and \vec{j}=\begin{bmatrix}0 \\ 1 \end{bmatrix}, or \vec{e}_1=\begin{bmatrix}1 \\ 0 \end{bmatrix} and \vec{e}_2=\begin{bmatrix}0 \\ 1 \end{bmatrix}, whichever you are used to. I'll use the latter.

    The area of the square spanned by \vec{e}_1 and \vec{e}_2 is clearly 1, so if we knew how the area of the parallelogram spanned by \vec{a} and \vec{b} was, we could then say how the map affects areas.

    So, what is the area of the parallelogram spanned by \vec{a} and \vec{b}?

    We can first find half its area, the triangle T, in the diagram below.

    Determinants: What is the motivation?-determinant-diagram.png
    From the diagram, T=a_2b_1 - T_1 - T_2 - T_3.

    Then, notice that
    T_1=\frac{1}{2}a_1a_2
    T_2=\frac{1}{2}(b_1-a_1)(a_2-b_2)
    T_3=\frac{1}{2}b_1b_2

    Do the algebra, you get T=\frac{1}{2}a_1b_2-\frac{1}{2}a_2b_1. Since the triangle's only half the area we want (the parallelogram), we double that quantity to get the determinant of a 2D matrix: a_1b_2-a_2b_1.

    The same reasoning applies to 3D determinants, just remember that the scalar triple product gives you the volume of the parallelopiped formed by three vectors.

    If you're asking for what it's good for, well, it's good for telling you whether a linear map changes area/volume. For example, shear maps don't change areas, so they have determinant 1. The determinant is also a signed quantity, so it also gives an indication of whether a reflection had taken place. If so, the determinant would be negative.

    Do you remember how to invert a 2x2 matrix? Then you might have seen this:

    \begin{bmatrix}a_1 & b_1 \\ a_2 & b_2\end{bmatrix}^{-1}=\frac{1}{a_1b_2-a_2b_1}\begin{bmatrix}b_2 & -b_1 \\ -a_2 & a_1\end{bmatrix}

    Never mind what's inside the matrix- do you see the determinant in the denominator there? If we applied the matrix onto some set of points, the area would've increased by a scale factor of a_1b_2 - a_2b_1, so if we want to reverse the transformation, we divide out by that same scale factor instead.

    But what if the scale factor is 0? What kind of vectors would have a parallelogram with 0 area? Well, two parallel vectors for a start, because there is no parallelogram in between the vectors! When that happens, a projection has occured, and information has been lost. It's like how when you take a photo, you lose depth information- everything has been projected onto the same plane so that depth information has been destroyed. If information has been lost, there is no way to restore it, so you cannot reverse the transformation, i.e, the matrix is not invertible.


    Going back to Gauss-Jordan elimination and solving a system of simultaneous equations. Suppose we have these equations:

    \begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}d_1 \\ d_2 \\ d_3\end{bmatrix}

    To solve the above, we want to isolate the x, y, z vector on its own. That's simple, and Gauss-Jordan essentially applies the matrix inverse to get

    \begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}^{-1} \begin{bmatrix}d_1 \\ d_2 \\ d_3\end{bmatrix}

    But that only works if our coefficient matrix is even invertible in the first place! If it is invertible, the determinant will be non-zero and everything's fine. If, on the other hand, the determinant is zero, a projection occured, information is lost and the matrix is not invertible. That means we do not have a unique solution.

    From the sound of it, this might be moving into new territory, so I'll stop there for now. Let us know if you're still stuck!
    Last edited by Tuufless; December 30th 2014 at 08:23 AM.
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