Description

Basic Properties of Determinants

1. The determinant of the identity matrix is $1$.

2. • Scaling one row by a factor $k$ multiplies the determinant by $k$.

     $\textbf{A} = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$

     Multiplying the first row by $k$

     $\text{det}\left(\begin{bmatrix} ka & kb \\ c & d\end{bmatrix}\right) = k \cdot \text{det}(A)$

   • Additivity

     $\text{det}\left(\begin{bmatrix}a + a' & b + b' \\ c & d\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix}a & b \\ c &d\end{bmatrix}\right) + \text{det}\left(\begin{bmatrix}a' & b' \\ c & d\end{bmatrix}\right)$

3. A row exchange reverses the sign of the determinant.

   $\text{det}\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right) = -\text{det}\left(\begin{bmatrix}c & d \\ a & b\end{bmatrix}\right)$

Using these properties allows you to derive many other properties of determinants!

Derived Properties of Determinants

1. If $\textbf{A}$ has two equal rows, $\text{det}(\textbf{A}) = 0$. Let $\textbf{A} = \begin{bmatrix}a & b \\ a & b\end{bmatrix}$. By basic property 3: $\text{det}\left(\begin{bmatrix}a & b \\ a & b\end{bmatrix}\right) = -\text{det}\left(\begin{bmatrix}a & b \\ a & b\end{bmatrix}\right)$ The only way this will hold true is if $\text{det}\left(\begin{bmatrix}a & b \\ a & b\end{bmatrix}\right) = 0$ From this, we conclude that if $\textbf{A}$ has two equal rows, $\text{det}(\textbf{A}) = 0$. If $\textbf{A}$ has two equal rows, this means that $\textbf{A}$ has linearly dependent rows. Consequently, we can conclude that if the $\text{det}(\textbf{A}) = 0$, the matrix has linearly dependent rows and is not invertible.

2. Adding or subtracting a row to or from another row does not change the determinant. Let $\textbf{A} = \left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)$ Solving for the determinant after multiplying the first row by t and adding it to the first row gives: $\text{det}\left(\begin{bmatrix}a & b \\ c + ta & d + tb\end{bmatrix}\right)$ By basic property 2 and derived property 1:

$\text{det}\left(\begin{bmatrix}a & b \\ c + ta & d + tb\end{bmatrix}\right) = \text{det}\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right) + \text{det}\left(\begin{bmatrix}a & b \\ ta & tb\end{bmatrix}\right)$ $=\text{det}\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right) + t \cdot \text{det}\left(\begin{bmatrix}a & b \\ a & b\end{bmatrix}\right)$ $\text{det}\left(\begin{bmatrix}a & b \\ a & b\end{bmatrix}\right) = 0$ Therefore: $\text{det}\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right) + t \cdot 0 =\text{det}\left(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\right)$ 3. The determinant of a matrix in upper triangular, or lower triangular, form is equal to the product of the values on the diagonal. $\text{det}\left(\begin{bmatrix}a & 0 \\ c & d\end{bmatrix}\right) = a \cdot d$ and $\text{det}\left(\begin{bmatrix}a & b \\ 0 & d\end{bmatrix}\right) = a \cdot d$ As a simpler case, let's prove that the determinant of a diagonal matrix is the product of the diagonal. Let $A = \left(\begin{bmatrix}a & 0 \\ 0 & d\end{bmatrix}\right)$ Using basic property 1 and 2a: $\text{det}(\textbf{A}) = a d \cdot \text{det}\left(\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)$ (Property 2) The determinant of the identity matrix is 1 (property 1). Therefore: $\text{det}(\textbf{A}) = a d \cdot 1 = a d$ This property can be extended to matrices in upper or lower triangular form using derived property 2. We can reduce a matrix in upper or lower triangular form to diagonal form by multiplying and adding rows to have 0's above all pivots.