If we take the definition of the determinant to be the amount by which areas/volumes are transformed by a linear transformation, a common definition, the determinant of a Matrix $A$ will be the area/volume contained in the parallelogram/parallelepiped formed by the matrix's vector components.

We can thus obtain an expression for the determinant by performing the elementary row operation of ading a multiple of one row to another until the matrix is diagonal, since this operation corresponds to a shear transformation the volume of the parallelepiped and thus the determinant will not change. Once the matrix is diagonal, the volume will simply be the diagonal elements multiplied together. This is easy to show for a $2\times2$ matrix: $$\begin{bmatrix} a&b \\ c&d \end {bmatrix}$$ This becomes $$\begin{bmatrix} a&0 \\ 0&d-\frac{bc}{a} \end {bmatrix}$$ So the determinant is $$\begin{vmatrix} a&b \\ c&d \end {vmatrix}=ad-bc$$ The same can be done for a $3\times3$ matrix $$\begin{bmatrix} a&b&c \\ d&e&f \\ g&h&i \end {bmatrix}$$ Which can be made into $$\begin{bmatrix} a&0&0 \\ 0&e-\frac{bd}{a}&f -\frac{cd}{a}\\ 0&h-\frac{bg}{a}&i-\frac{cg}{a} \end {bmatrix}$$ So the determinant will be $$\begin{vmatrix} a&b&c \\ d&e&f \\ g&h&i \end {vmatrix}= a\begin{vmatrix} e-\frac{bd}{a}&f -\frac{cd}{a}\\ h-\frac{bg}{a}&i-\frac{cg}{a} \end {vmatrix}$$ Which can be rearranged to $$\begin{vmatrix} a&b&c \\ d&e&f \\ g&h&i \end {vmatrix} = a\begin{vmatrix} e&f\\ h&i \end {vmatrix}-b \begin{vmatrix} d&f \\ g&i \end {vmatrix}+c \begin{vmatrix} d&e \\ g&h \end {vmatrix}$$ My question is, how can this be generalized to higher dimensional matrices using the same method?

  • $\begingroup$ Look up any linear algebra text on the computation of a determinant by Gaussian elimination (for this is what you've been doing in the $2 \times 2$ and $3 \times 3$ cases). $\endgroup$ – darij grinberg Sep 8 '18 at 23:44

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