# 3x3 matrix operations intuition help

So I understand the intuition of taking the determinant of a 2x2 matrix, but what is the intuition for taking the determinant of 3x3, matrix? It makes zero intuitive sense just looking at it. Also, when finding the inverse of a matrix, why do we need to find the cofactor matrix and the adjugate matrix and transpose the cofactor matrix, and what is the point of the checkerboard matrix with + and - signs?

• There are many "intuitions" behind a determinant. What is your understanding for 2 by 2 matrix? – Todor Markov Jan 29 '19 at 14:13
• Determinant intuition: youtube.com/watch?v=Ip3X9LOh2dk – MCCCS Jan 29 '19 at 14:19
• I’ve seen that video, but what about 3x3 matrices? – Jon due Jan 29 '19 at 14:34
• One "intuition" is as a scaling factor for linear transformations. Suppose the 2D matrix $T$ represents a L.T., and suppose $D=\{(x,y)\mid x^2+y^2<r^2\}$ is the unit disc with area $a=\pi r^2$. Transforming the disc by $T$, the area becomes $a'=\pi r^2\det T$. In the 3D case you might have the ball $E=\{(x,y,z)\mid x^2+y^2+z^2<r^2\}$, and a 3D linear transformation matrix $U$. The ball $E$ has volume $a=4\pi r^3/3$. After transforming the ball by the transformation $U$ it would have volume $a'=4\pi r^3/3\det U$. Perhaps this isn't the kind of intuition you're looking for though... – Pixel Jan 29 '19 at 14:37

Suppose we have a matrix $$A=(a_{ij})_{ij}$$. Lets denote the determinant of $$A$$ once we remove the $$i$$th row and the $$j$$th column as $$A_{i,j}$$ and lets denote $$(-1)^{i+j}\cdot A_{i,j}$$ as $$C_{i,j}$$ (the $$i,j$$ cofactor of $$A$$). Then, by definition, the determinant of a matrix of dimension 1x1 is the number itself and for any other dimension of $$A$$ is:

$$|A| = a_{11}C_{11} + a_{12}C_{12} + \dots + a_{1n}C_{1n}$$

(By Laplace extension's Theorem you can use any row not necessarily the first one)

that means that

$$\begin{vmatrix}a&b\\c&d\end{vmatrix} = a \begin{vmatrix}d\end{vmatrix} - b\begin{vmatrix}c\end{vmatrix} = ad-bc$$

And in the case of a 3x3 matrix

$$\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix} = a\begin{vmatrix}e&f\\h&i\end{vmatrix}-b\begin{vmatrix}d&f\\g&i\end{vmatrix}+c\begin{vmatrix}d&e\\g&h\end{vmatrix} = aei-afh-bdi+bgf+cdh-cge$$

So even though at first glance the determinant of a 3x3 matrix it may seem a non intuitive result, it actually is a result of computing determinants 2x2.

• But why do we do that? Why eliminate certain rows and columns? Why use the first 3 entries as scalars? – Jon due Jan 29 '19 at 14:34
• In linear algebra most things are from definition. In this case, you define determinant to be that because you realize that the number you get by doing this computation has some relevance: it can tell whether a matrix in invertible or not, whether a system is inconsistent, etc. – maxbp Jan 29 '19 at 14:40
• It is not only that, don't get me wrong, the determinant of the standard matrix of a lineal map represents the stretchment of the vectors once the map is applied. But for example, the cofactor matrix is defined that way so you can express the determinant, the adjoint matrix etcetera, in a more compact way – maxbp Jan 29 '19 at 14:45
• What about the checkerboard matrix with the + and - signs? Thank you for your help so far. – Jon due Jan 29 '19 at 17:04
• this is done to represent the sign of the cofactor, if you look at the definition of cofactor (2n line of the post) you will realize that there is the (-1)^{i+j}. Meaning that -1 appear when i+j is odd giving you that pattern +-+-+-+ ... And if you want to know the reason why we put the negative signs in the cofactor well, that is again by definition. If you do the same with with all positive sign it gives you some irrelevant information, whereas with negative values in the odd positions gives you something interesting, worth to put it a name. You're welcome, I hope I was useful! – maxbp Jan 29 '19 at 17:10