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# 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.