Determinants

Definition: Cofactor Matrix

Let $A\in F^{n\times n}$ be a square matrix.

The cofactor matrix of an element $a_{ij}$ of $A$ is the matrix $A_{ij}\in F^{(n-1)\times (n-1)}$ obtained by removing the $i$-th row and the $j$-th column of $A$ and then sticking the rest of its rows and columns together.

EXAMPLE

$$\left[\begin{array}{ccc}6& 7& 8\\ 3& 2& 1\\ 7& 6& 5\end{array}\right]$$

The cofactor matrix of the element $a_{1,1}=6$ is

$$\left[\begin{array}{cc}2& 1\\ 6& 5\end{array}\right]$$

The cofactor matrix of the element $a_{1,2}=7$ is

$$\left[\begin{array}{cc}3& 1\\ 7& 5\end{array}\right]$$

The cofactor matrix of the element $a_{1,3}=8$ is

$$\left[\begin{array}{cc}3& 2\\ 7& 6\end{array}\right]$$

Definition: Determinant

The determinant of a square matrix $A\in F^{n\times n}$ is an element in $F$ which is calculated recursively from the coefficients of $A$ and their cofactor matrices as follows:

$$\det (A)=\{\begin{array}{l}{a}_{11},\text{ if }n=1\\ \sum _{i=1}^n(-1{)}^{1+j}{a}_{1j}\det (A_{1j}),\text{ if }n>1\end{array}$$

NOTATION

$$|A|\det (A)$$

Theorem: Distributivity of the Determinant

The Determinants is distributive over matrix products:

$$\det (AB)=\det (A)\det (B)=\det (B)\det (A)=\det (BA)$$

PROOF

TODO

Theorem: Determinant of the Transpose

The Determinants of the transpose of a $A$ is the same as the Determinants of $A$.

$$\det (A^{\mathsf{T}})=\det (A)$$

PROOF

TODO

Theorem: Determinant of the Inverse

If $A$ is invertible, then the Determinants of $A^{-1}$ is the reciprocal of $A$‘s Determinants:

$$\det (A^{-1})=\frac{1}{\det (A)}$$

PROOF

TODO

Theorem: Determinant of Scalar Multiplication

For the Determinants of every Square Matrices $A\in F^{n\times n}$ and every $\alpha \in F$:

$$\det (\alpha A)={\alpha }^n\det (A)$$

PROOF

TODO

Theorem: Effects of Row and Column Operations on Determinants

For any square matrix $A\in F^{n\times n}$:

• Swapping two rows or two columns changes the algebraic sign of $A$‘s determinant.

• Multiplying a single row or a single column by $k\in F$ results in $A$‘s determinant being multiplied by $k$.

• Adding a non-zero multiple of one row or column to another row or column has no effect on $A$‘s determinant.

PROOF

TODO

Calculating Determinants

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 determinants of the cofactor matrices 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 Matrices , 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. 3. For the first matrix:

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

4. For the second matrix:

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

5. For the third matrix:

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

6. 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.