The determinant of a matrix represents the *scaling factor* applied by the transformation described by the matrix. Geometrically, it measures how the transformation stretches or compresses areas ("2D"), volumes ("3D") or hyperplanes. For example, consider a transformation of the unit square in two dimensions: $ \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \to \begin{bmatrix}a & b \\ c & d \end{bmatrix} $ The determinant for a $2 \times 2$ matrix is computes as: $ A= \begin{bmatrix} a & b\\ c & d \end{bmatrix} \implies \det (A)=ad-bc $ This value corresponds to the area of the parallelogram formed by the transformed unit square. ![[matrix-determinant.png|center|300]] ## Derivation The determinant can be derived from the [[Inverse Matrix#^81f95f|matrix inverse property]] $A \cdot A^{-1}=\mathbf I$, where $\mathbf I$ is the identity matrix. To get from $A$ to the identity $\mathbf I$, we take the inverse directions and scale them according to the determinant. $ A= \begin{bmatrix} a&b \\ c&d \end{bmatrix}, \qquad A^{-1}=\frac{1}{\det (A)} \begin{bmatrix} d&-b \\ -c&a \end{bmatrix}$ Compute $A \cdot A^{-1}$: $ A \cdot A^{-1} =\begin{bmatrix} a&b \\ c&d \end{bmatrix} \cdot \frac{1}{\det (A)} \begin{bmatrix} d&-b \\ -c&a \end{bmatrix} $ Resulting in: $ A \cdot A^{-1} = \frac{1}{\det (A)} \cdot \begin{bmatrix} ad-bc & 0 \\ 0 & ab-bc \end{bmatrix} $ Since this expression needs equal the identity matrix, we need to normalize diagonal elements to $1$. $ \frac{1}{\det (A)}*(ad-bc)=1 \quad \implies \quad \det(A) = ad-bc$ >[!note:] >Finding determinants for bigger than $2 \times 2$ matrices is more complicated, but can be done by software programs. ## Linear Independence The determinant provides insights into the linear independence of the matrix's column vectors: - $\det(A)=0$: Vectors are *linearly dependent*, meaning one vector is a linear combination of others. Such a matrix collapses a dimension during transformation and does not have an inverse. - $\det(A) \neq 0$: Vectors are [[Linearly Independent Vectors|linearly independent]], and the transformation can be reversed using the inverse matrix. Example: Consider a degenerate matrix $ a_1= \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad a_2= \begin{bmatrix} 2 \\ 2 \end{bmatrix}, \quad A= \begin{bmatrix} 1 & 2 \\ 1& 2 \end{bmatrix} $ Compute the determinant: $ \det(A)=(1_2)-(1_2)=0 $ Since the determinant is $0$, the column vectors $a_1$​ and $a_2$​ are linearly dependent, and $A$ is not invertible.