Calculating the Determinant

Algorithm: Calculating the Determinant of a Matrix

To calculate the determinant of a square matrix $A\in F^{n\times n}$:

1. Go through the entries in the first row of $A$ one by one. Multiply the $k$-th entry $A_{1,k}$ with the determinant of its cofactor matrix $A_{1,k}$ and alternate the algebraic sign each time - if $k$ is even, place a minus sign before the result. Calculating the determinant of the cofactor matrix involves the same process recursively, until a $3\times 3$ or $2\times 2$-matrix is obtained, at which point one can use the theorems for those.

2. The sum of all results from Step 1 is the determinant of $A$.

Tips

1. Search for a row or column with many rows and exchange it with the first one.
2. If the first column contains many zeros, calculate the determinant of $A$‘s transpose, since they are the same.

EXAMPLE

$$A=\left[\begin{array}{cccc}1& 2& 3& 4\\ 5& 6& 7& 8\\ 4& 3& 2& 1\\ 8& 7& 6& 5\end{array}\right]$$

According to the algorithm

$$\det (A)=1\left|\begin{array}{ccc}6& 7& 8\\ 3& 2& 1\\ 7& 6& 5\end{array}\right|-2\left|\begin{array}{ccc}5& 7& 8\\ 4& 2& 1\\ 8& 6& 5\end{array}\right|+3\left|\begin{array}{ccc}5& 6& 8\\ 4& 3& 1\\ 8& 7& 5\end{array}\right|-4\left|\begin{array}{ccc}5& 6& 7\\ 4& 3& 2\\ 8& 7& 6\end{array}\right|$$

We must now calculate the determinants of the $3\times 3$-matrices.

1. For the first matrix:

$$\left|\begin{array}{ccc}6& 7& 8\\ 3& 2& 1\\ 7& 6& 5\end{array}\right|=6\left|\begin{array}{cc}2& 1\\ 6& 5\end{array}\right|-7\left|\begin{array}{cc}3& 1\\ 7& 5\end{array}\right|+8\left|\begin{array}{cc}3& 2\\ 7& 6\end{array}\right|$$ $$=6(2\cdot 5-1\cdot 6)-7(3\cdot 5-1\cdot 7)+8(3\cdot 6-2\cdot 7)$$ $$=6(10-6)-7(15-7)+8(18-14)$$ $$=6\cdot 4-7\cdot 8+8\cdot 4$$ $$=24-56+32=0$$

2. For the second matrix:

$$\left|\begin{array}{ccc}5& 7& 8\\ 4& 2& 1\\ 8& 6& 5\end{array}\right|=5\left|\begin{array}{cc}2& 1\\ 6& 5\end{array}\right|-7\left|\begin{array}{cc}4& 1\\ 8& 5\end{array}\right|+8\left|\begin{array}{cc}4& 2\\ 8& 6\end{array}\right|$$ $$=5(2\cdot 5-1\cdot 6)-7(4\cdot 5-1\cdot 8)+8(4\cdot 6-2\cdot 8)$$ $$=5(10-6)-7(20-8)+8(24-16)$$ $$=5\cdot 4-7\cdot 12+8\cdot 8$$ $$=20-84+64=0$$

3. For the third matrix:

$$\left|\begin{array}{ccc}5& 6& 8\\ 4& 3& 1\\ 8& 7& 5\end{array}\right|=5\left|\begin{array}{cc}3& 1\\ 7& 5\end{array}\right|-6\left|\begin{array}{cc}4& 1\\ 8& 5\end{array}\right|+8\left|\begin{array}{cc}4& 3\\ 8& 7\end{array}\right|$$ $$=5(3\cdot 5-1\cdot 7)-6(4\cdot 5-1\cdot 8)+8(4\cdot 7-3\cdot 8)$$ $$=5(15-7)-6(20-8)+8(28-24)$$ $$=5\cdot 8-6\cdot 12+8\cdot 4$$ $$=40-72+32=0$$

4. For the fourth matrix:

$$\left|\begin{array}{ccc}5& 6& 7\\ 4& 3& 2\\ 8& 7& 6\end{array}\right|=5\left|\begin{array}{cc}3& 2\\ 7& 6\end{array}\right|-6\left|\begin{array}{cc}4& 2\\ 8& 6\end{array}\right|+7\left|\begin{array}{cc}4& 3\\ 8& 7\end{array}\right|$$ $$=5(3\cdot 6-2\cdot 7)-6(4\cdot 6-2\cdot 8)+7(4\cdot 7-3\cdot 8)$$ $$=5(18-14)-6(24-16)+7(28-24)$$ $$=5\cdot 4-6\cdot 8+7\cdot 4$$ $$=20-48+28=0$$

We see that all determinants are $0$ and so

$$\det (A)=1\cdot 0-2\cdot 0+3\cdot 0-4\cdot 0=0$$

Theorem: Determinant of a $2\times 2$-Matrix

The determinant of every $2\times 2$-matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is given by

$$\det (A)=ad-bc$$

PROOF

By definition

$$\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=a\det (\left[\begin{array}{c}d\end{array}\right])-b\det (\left[\begin{array}{c}c\end{array}\right])=ad-bc$$

Theorem: Determinant of a $3\times 3$-Matrix

The determinant of every $3\times 3$-matrix $A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$ is given by

$$\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=a(ei-fh)-b(di-fg)+c(dh-ge)$$

PROOF

By definition

$$\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=a\left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b\left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c\left|\begin{array}{cc}d& e\\ g& h\end{array}\right|$$

The $2\times 2$-determinants can be calculated with the previous theorem.