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EDIT: corrected a typo in the subject.

A while ago I have learned the mechanics of finding the determinant of an matrix $A_{nxn}$. I am attaching a screen shot with an example, which includes steps using cofactors. But my problem is the following: as time went by I realized that the only thing I know is this mechanical (technical) approach to solution, but I do not understand:

(1) the logic of each $A_{ij}$ determinant used to compute the overall determinant of the $A_{n\times n}$; what it is; why it is computed this way, etc. I mean that I don't understand it's meaning and it's purpose, like for example we all understand that if we say "a fire" we know what it does technically, but also we understand what it means, what it can do and why it does what it does;

(2) the same goes to understanding, or rather non-understanding, the logic and the meaning of computing the overall $A_{n\times n}$ determinant using a multiplication of $a_{ij}$ values by $A_{ij}$ determinant.

I am not sure I managed to explain myself clearly enough, but, again, I don't mean the technicalities, as they seem to be clear, more or less, but the meaning.

Here is the picture with technically correct example (there are other ways also, for example expanding over the second row, but I chose this option) of a $A_{3\times 3}$ matrix and a general $A_{n\times n}$ formula:

$$ A_{3\times 3} = \begin{bmatrix} 3 & 1 & 2 \\ 1 & -1 & 5 \\ 2 & 1 & 4 \end{bmatrix} $$

$$det (A_{nxn}) = 3 det(A_{1x1}) - 1 det(A_{1x2}) + 2 det(A_{1x3}) = $$ $$ 3 (-1 * 4 - 5) - (4 - 10) + 2 (1 + 2) = -15 $$

[EDIT: corrected typos]

And the general formula:

$$ a_{11} (-1) ^ {1+1} det(A_{1x1}) - a_{12} (-1) ^ {1+2} det(A_{1x2}) +...+ a_{in}(-1)^{1+n} det(A_{ixn})$$

where each $(-1)^{1+n} det(A_{i\times n})$ is a cofactor of the matrix.

I will be very grateful for your help!

Thank you very much!

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  • $\begingroup$ Your worked out example is incorrect, the determinant is $-15$. If, as it appears, you expanded over the first row, then it should be $3(-1 \times 4 - 5) - (4 - 10) + 2(1 + 2) = -15$. $\endgroup$ – Lee Mosher Jun 25 '17 at 15:01
  • $\begingroup$ The site supports posting mathematical expressions using $\LaTeX$ syntax. The introduction has links to more advance features in the home-grown tutorial and other resources. $\endgroup$ – hardmath Jun 25 '17 at 15:12
  • $\begingroup$ @LeeMosher Thank you. Sorry about these typos. I corrected them. Though my questions still hold. $\endgroup$ – Vitale Jun 25 '17 at 15:34
  • $\begingroup$ @hardmath Thank you. Yes, I know these links, but I couldn't find a way to draw a matrix there. I tried to use 'big brackets' but those didn't work ([QUOTE] like these ones $\Biggl(\biggl(\Bigl(\bigl((x)\bigr)\Bigr)\biggr)\Biggr)$ gives (((((x))))) [\QUOTE]) Ok, here they worked, in this comment, but I couldn't get them surround the matrix. $\endgroup$ – Vitale Jun 25 '17 at 15:37
  • $\begingroup$ In the title, by "marries" did you mean "matrix", or something else? $\endgroup$ – Andrés E. Caicedo Jun 25 '17 at 15:37
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The determinant of an $(n\times n)$-matrix $A=[A_{ik}]_{1\leq i\leq n, \ 1\leq k\leq n}$ is a very complicated polynomial function of the $n^2$ entries $A_{ik}$ of $A$. This polynomial consists of $n!$ terms formed in a certain symmetric way. It does not make sense to write out all these terms when $n\geq4$ or so. Now the scheme you have learned is a recursive principle: It tells you how to compute (or "build up") an $(n\times n)$-determinant from $(n-1)\times(n-1)$- determinants.

This does not yet answer your question of why we should look at determinants at all. If the entries $A_{ik}$ are the prices of $10$ different commodities in $10$ different countries, arranged in a matrix ("spreadsheet") $A$, then nobody in his right mind would care about ${\rm det}(A)$. But more often than not a matrix $A$ is set up with some other, e.g., physical or geometric, application in mind. In such cases it typically is the matrix of a linear map $A:\>V\to V$, whereby $V$ is an $n$-dimensional vector space with a chosen basis. In this case the number ${\rm det}(A)$ has, among other things, an interesting geometric meaning: The map $A$ deforms any set $\Omega\subset V$ affinely to a new set $A(\Omega)$, and one has the formula $${\rm vol}\bigl(A(\Omega)\bigr)=|{\rm det}(A)|\cdot{\rm vol}(\Omega)\ ,$$ valid for all measurable sets $\Omega$. This is indeed a miracle, and is the secret behind the appearance of the Jacobian determinant in the substitution formula for multiple integrals.

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  • $\begingroup$ This is fascinating. I am very thankful to you for this explanation. To be honest, I didn't understand your example (I think I slightly understand the idea, but not the mechanics and the syntax), because I don't know linear algebra (vectors), as well as set theory, yet. I will try to read more on this to understand this very interesting example. $\endgroup$ – Vitale Jul 2 '17 at 12:58

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