17.1: Introduction to Determinants

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urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py', 
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For a detailed overview of determinants I would recommend reviewing Chapter D pg 340-366 of the Beezer text.

The determinant is a function that takes a (\(n \times n\)) square matrix as an input and produces a scalar as an output. Determinants have been studied quite extensively and have many interesting properties. However, determinants are “computationally expensive” as the size of your matrix (\(n\)) gets bigger. This limitation makes them impractical for many real world problems.

The determinant of a \(2 \times 2\) matrix can be calculated as follows:

\[\begin{split}
det \left(
\left[
\begin{matrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{matrix}
\right]
\right)
= a_{11}a_{22} - a_{12}a_{21}
\end{split} \nonumber \]

\[\begin{split}
\left[
\begin{matrix}
3 & -2 \\
1 & 2
\end{matrix}
\right]
\end{split} \nonumber \]

Calculating the determinant of a larger matrix is a “recursive” problem which involves combining the determinants of smaller and smaller sub-matrices until you have a \(2 \times 2\) matrix which is then calculated using the above formula. Here is some Pseudocode to calculate a determinant. To simplify the example the code assumes there is a matrix function deleterow which will remove the \(x\)th row from a matrix (always the first row in this example) and deletecol will remove the \(x\)th column from a matrix. When used together (as shown below) they will take an \(n \times n\) matrix and turn it into a \((n−1) \times (n−1)\) matrix.

function determinant(A, n)
   det = 0
   if (n == 1)
      det = matrix[1,1]
   else if (n == 2)
      det = matrix[1,1] * matrix[2,2] - matrix[1,2] * matrix[2,1]
   else 
      for x from 1 to n
          submatrix = deleterow(matrix, 1)
          submatrix = deletecol(submatrix, x)
          det = det + (x+1)**(-1) * matrix[1,x] * determinant(submatrix, n-1)
      next x
   endif
   
   return det

Notice that the combination of the determinants of the submatrixes is not a simple sum. The combination is adding the submatrices corresponding to the odd columns (1,3,5, etc) and subtracting the submatrices corresponding to the even columns (2,4,6, etc.). This may become clearer if we look at a simple \(3 \times 3\) example (Let \(\left| A \right|\) be a simplified syntax for writing the determinant of \(A\)):

\[\begin{split}
A = \left[
\begin{matrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{matrix}
\right] \end{split} \nonumber \]

\[\begin{split}
|A|=
a_{11} \left|
\begin{matrix}
\square & \square & \square \\
\square & a_{22} & a_{23} \\
\square & a_{32} & a_{33}
\end{matrix}
\right|
-
a_{12}\left|
\begin{matrix}
\square & \square & \square \\
a_{21} & \square & a_{23} \\
a_{31} & \square & a_{33}
\end{matrix}
\right|
+
a_{13} \left|
\begin{matrix}
\square & \square & \square \\
a_{21} & a_{22} & \square \\
a_{31} & a_{32} & \square
\end{matrix}
\right|
\end{split} \nonumber \]

\[\begin{split}
|A|
=
a_{11}\left|
\begin{matrix}
a_{22} & a_{23} \\
a_{32} & a_{33}
\end{matrix}
\right|
-
a_{12}\left|
\begin{matrix}
a_{21} & a_{23} \\
a_{31} & a_{33}
\end{matrix}
\right|
+
a_{13}
\left|
\begin{matrix}
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{matrix}
\right|
\end{split} \nonumber \]

\[
|A| =
a_{11}(a_{22}a_{33} - a_{23}a_{32})
-
a_{12}(a_{21}a_{33} - a_{23}a_{31})
+
a_{13}(a_{21}a_{32} - a_{22}a_{31}) \nonumber \]

\[\begin{split}
\left[
\begin{matrix}
1 & 2 & -3 \\
5 & 0 & 6 \\
7 & 1 & -4
\end{matrix}
\right]
\end{split} \nonumber \]

\[\begin{split}
\left[
\begin{matrix}
2 & 0 & 1 & -5 \\
8 & -1 & 2 & 1 \\
4 & -3 & -5 & 0 \\
1 & 4 & 8 & 2
\end{matrix}
\right]
\end{split} \nonumber \]

#Put your answer here

from answercheck import checkanswer

checkanswer.float(det,'49afb719e0cd46f74578ebf335290f81');