Determinants
Determinants
We need to define
determinants and list a few of their principal properties.
- For $2 \times 2$ matrices,
$$\textrm{det} \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right]
= a d - b c.$$
- For $3 \times 3$ matrices,
$$\begin{align}
\textrm{det} \left[ \begin{array}{ccc} a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \\
\end{array} \right] & =
a_1 \cdot \textrm{det} \left[ \begin{array}{cc}
b_{2} & b_{3} \\c_{2} & c_{3} \\ \end{array} \right]
-a_2 \cdot \textrm{det} \left[ \begin{array}{cc}
b_{1} & b_{3} \\c_{1} & c_{3} \\ \end{array} \right]
+a_3 \cdot \textrm{det} \left[ \begin{array}{cc}
b_{1} & b_{2} \\c_{1} & c_{2} \\ \end{array} \right] \\
& = a_1 (b_2 c_3 - b_3 c_2) - a_2 (b_1 c_3 - b_3 c_1) + a_3 (b_1 c_2 - b_2 c_1) \\
\end{align}
$$
For example,
$$\textrm{det} \left[ \begin{array}{cc} 4 & 2 \\ 3 & 1 \\ \end{array} \right]
\ = \ 4 \cdot 1 \, - \, 2 \cdot 3 \ = \ -2 \, ,$$
and
$$\textrm{det} \left[ \begin{array}{ccc} 1 & 2 & -3 \\ 4 & 1 & 1 \\
1 & 2 & -1 \\ \end{array} \right]
\ = \ 1 \cdot \textrm{det} \left[ \begin{array}{cc} 1 & 1 \\ 2 & -1 \\
\end{array} \right]
- 2 \cdot \textrm{det} \left[ \begin{array}{cc} 4 & 1 \\ 1 & -1 \\
\end{array} \right] -3 \cdot \textrm{det} \left[ \begin{array}{cc}
4 & 1 \\ 1 & 2 \\ \end{array} \right] \ = \ -3
-2(-5) -3(7) \ = \ -14\,. $$
The formula for the general determinant depends on submatrices.
Let $A = \left[ a_{i,\,j} \right] \in M_n(\mathbb{R}).$ For any pair
$(i,j)$ with $1 \leq i , j \leq n$ we define the Submatrix
$$A_{i,\,j} \in M_{n-1}(\mathbb{R})$$
be deleting the $i^{th}$ row and $j^{th}$ column of $A.$
For example, if
$$A = \left[ \begin{array}{ccc} 1 & 2 & -3 \\ 4 & 1 & 1 \\
1 & 2 & -1 \\ \end{array} \right]$$
then
$$A_{1,1} = \left[ \begin{array}{cc} 1 & 1 \\ 2 & -1 \\
\end{array} \right]$$
and
$$A_{3,2} = \left[ \begin{array}{cc} 1 & -3 \\ 4 & 1 \\
\end{array} \right].$$
For $A = \left[ a_{i,\,j} \right] \in M_n(\mathbb{R}),$ we define
the Determinant of $A$ as
$$
\textrm{det} (A) \ = \ a_{1,1} \textrm{det} (A_{1,1}) - a_{1,2}
\textrm{det} (A_{1,2}) + \ldots +
(-1)^{n-1}a_{1,\, n} \textrm{det} (A_{1,\, n}).$$
Let's see how this gives the determinant of a $4 \times 4:$
$$\begin{align}
\textrm{det} \left[ \begin{array}{cccc} 2 & 7 & -1 & 5 \\
-3 & 4 & 2 & 0 \\ 3 & 1 & -2 & 4 \\ 0 & 2 & 5 & 3 \\
\end{array} \right] \ = \ &
2 \cdot \textrm{det} \left[ \begin{array}{ccc} 4 & 2 & 0 \\
1 & -2 & 4 \\ 2 & 5 & 3 \\ \end{array} \right]
- 7 \cdot \textrm{det} \left[ \begin{array}{ccc} -3 & 2 & 0 \\
3 & -2 & 4 \\ 0 & 5 & 3 \\ \end{array} \right] \\
& - 1 \cdot \textrm{det} \left[ \begin{array}{ccc} -3 & 4 & 0 \\
3 & 1 & 4 \\ 0 & 2 & 3 \\ \end{array} \right]
-5 \cdot \textrm{det} \left[ \begin{array}{ccc} -3 & 4 & 2 \\
3 & 1 & -2 \\ 0 & 2 & 5 \\ \end{array} \right]. \\
\end{align}$$
Let $A , B \in M_n(\mathbb{R}).$ Then
- $\textrm{det} (AB) = \textrm{det}(A) \textrm{det}(B).$
- $\textrm{det} \left( A^{-1} \right) =
\displaystyle\frac{1}{\textrm{det} (A)}.$
Finally, the most important fact about determinants for us is
that a matrix is nonsingular if and only if its determinant is nonzero.
We will add this to our list of statements equivalent to a matrix
being nonsingular.
Theorem: For $A \in M_n(\mathbb{R}),$ the following are
equivalent:
- $A$ is invertible, i.e., there exists a matrix
$A^{-1} \in M_n(\mathbb{R})$ such that $A A^{-1} = A^{-1} A = I_n.$
- The matrix equation $A {\bf x} \, = \, {\bf b}$ has
a unique solution for every ${\bf b} \in {\mathbb R}^n.$
- The matrix equation $A {\bf x} \, = \, {\bf 0}$ has
only the trivial solution.
- The reduced row echelon form of $A$ is $I_n,$ i.e.,
$\textrm{RREF}(A) = I_n.$
- The determinant of $A$ is nonzero: $\textrm{det}(A) \neq 0.$