Editing Invertible matrix (section)

=== Reciprocal basis vectors method ===
{{Main|Reciprocal basis}}
Given an {{math|''n'' × ''n''}} square matrix <math>\mathbf{X} = \left[ x^{ij} \right] </math>, <math> 1 \leq i,j \leq n </math>, with {{mvar|n}} rows interpreted as {{mvar|n}} vectors <math>\mathbf{x}_{i} = x^{ij} \mathbf{e}_{j}</math> ([[Einstein summation]] assumed) where the <math>\mathbf{e}_{j}</math> are a standard [[orthonormal basis]] of [[Euclidean space]] <math>\mathbb{R}^{n}</math> (<math>\mathbf{e}_{i} = \mathbf{e}^{i}, \mathbf{e}_{i} \cdot \mathbf{e}^{j} = \delta_i^j</math>), then using [[Clifford algebra]] (or [[geometric algebra]]) we compute the reciprocal (sometimes called [[Geometric algebra#Dual basis|dual]]) column vectors:
:<math>\mathbf{x}^{i} = x_{ji} \mathbf{e}^{j} = (-1)^{i-1} (\mathbf{x}_{1} \wedge\cdots\wedge ()_{i} \wedge\cdots\wedge\mathbf{x}_{n}) \cdot (\mathbf{x}_{1} \wedge\ \mathbf{x}_{2} \wedge\cdots\wedge\mathbf{x}_{n})^{-1} </math> 
as the columns of the inverse matrix <math>\mathbf{X}^{-1} = [x_{ji}].</math> Note that, the place "<math>()_{i}</math>" indicates that "<math>\mathbf{x}_{i}</math>" is removed from that place in the above expression for <math>\mathbf{x}^{i}</math>. We then have <math>\mathbf{X}\mathbf{X}^{-1} = \left[ \mathbf{x}_{i} \cdot \mathbf{x}^{j} \right] = \left[ \delta_{i}^{j} \right] = \mathbf{I}_{n} </math>, where <math>\delta_{i}^{j}</math> is the [[Kronecker delta]]. We also have <math>\mathbf{X}^{-1}\mathbf{X} = \left[\left(\mathbf{e}_{i}\cdot\mathbf{x}^{k}\right)\left(\mathbf{e}^{j}\cdot\mathbf{x}_{k}\right)\right] = \left[\mathbf{e}_{i}\cdot\mathbf{e}^{j}\right] = \left[\delta_{i}^{j}\right] = \mathbf{I}_{n}</math>, as required. If the vectors <math>\mathbf{x}_{i}</math> are not linearly independent, then <math>(\mathbf{x}_{1} \wedge \mathbf{x}_{2} \wedge\cdots\wedge\mathbf{x}_{n}) = 0</math> and the matrix <math>\mathbf{X}</math> is not invertible (has no inverse).