[BUG] problem in math_linear_algebra.ipynb

[xmg333]

Dear Author,

I'm reading the Determinant part of the math notebook,
And I suppose the row and column of the matix M didn't marked correctly? Maybe they should be inverted?

## Determinant
The determinant of a square matrix $M$, noted $\det(M)$ or $\det M$ or $|M|$ is a value that can be calculated from its elements $(M_{i,j})$ using various equivalent methods. One of the simplest methods is this recursive approach:

$|M| = M_{1,1}\times|M^{(1,1)}| - M_{2,1}\times|M^{(2,1)}| + M_{3,1}\times|M^{(3,1)}| - M_{4,1}\times|M^{(4,1)}| + \cdots ± M_{n,1}\times|M^{(n,1)}|$

* Where $M^{(i,j)}$ is the matrix $M$ without row $i$ and column $j$.

For example, let's calculate the determinant of the following $3 \times 3$ matrix:

$M = \begin{bmatrix}
  1 & 2 & 3 \\
  4 & 5 & 6 \\
  7 & 8 & 0
\end{bmatrix}$

Using the method above, we get:

$|M| = 1 \times \left | \begin{bmatrix} 5 & 6 \\ 8 & 0 \end{bmatrix} \right |
     - 2 \times \left | \begin{bmatrix} 4 & 6 \\ 7 & 0 \end{bmatrix} \right |
     + 3 \times \left | \begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix} \right |$

Now we need to compute the determinant of each of these $2 \times 2$ matrices (these determinants are called **minors**):

$\left | \begin{bmatrix} 5 & 6 \\ 8 & 0 \end{bmatrix} \right | = 5 \times 0 - 6 \times 8 = -48$

$\left | \begin{bmatrix} 4 & 6 \\ 7 & 0 \end{bmatrix} \right | = 4 \times 0 - 6 \times 7 = -42$

$\left | \begin{bmatrix} 4 & 5 \\ 7 & 8 \end{bmatrix} \right | = 4 \times 8 - 5 \times 7 = -3$

Now we can calculate the final result:

$|M| = 1 \times (-48) - 2 \times (-42) + 3 \times (-3) = 27$

[ageron]

Oh good catch, thanks! I inverted columns and rows in the notation. I just fixed this, it now reads:

[ageron]

I should add that the way it was written before actually gives the correct result as well (it's equivalent), but it didn't match the example, so it was confusing.