# Geometric intuition behind determinant properties

Currently I am working through Linear Algebra done wrong, in which the determinant is derived from a few assumed properties of higher dimensional volume of a parallelepiped. Let $D(x_1,x_2,...,x_n)=Det(A)$, where $A$ is the matrix with columns $x_1,x_2...x_n$ Then one of the properties given as a one of the core geometric principles that the determinant follows is $$D(x_1,...,x_j+ \alpha x_k,...,x_k,...,x_n)= D(x_1,...,x_j,...,x_k,...,x_n)$$ Whilst i have been able to verify this in $2$ dimensions, i have found myself unable to geometrically explain this fact in $3$ dimensions or higher. Any help with this would be greatly appreciated.

• – Mauve Jan 16 '18 at 4:02

## 1 Answer

Shears do not change the volume. Just as a parallelogram has the same area as a rectangle with same height, a shear $n$-dimensional parallelopiped has the same measure as one with the same $(n-1)$-dimensional base and height.

Your left-hand side $n$-dimensional parallelopiped has sheared the height in the $j$th dimension into a direction (the $x_k$ direction) parallel to its $(n-1)$-dimensional base, spanned by $x_1,\dotsc,x_{j-1},x_{j+1},\dotsc,x_n$, relative to the parallelopiped on the right-hand side. Since this shear is orthogonal to $x_j$, the height in the $x_j$ dimension remains unchanged. As does the size of the $(n-1)$-dimensional base. And therefore also the total $n$-dimensional volume.

• Thank you, the orthogonality was the piece of the puzzle that i was missing. – B.Martin Jan 16 '18 at 5:39