An n x n matrix, times its MatrixInverse, equals the IdentityMatrix?.
**A** is multiplied by an n x n identity matrix **I**, the answer is **A**. (Just like when a scalar *a* is multiplied by 1, the answer is *a*.)
Some square matrices do not have inverses.
The MatrixInverse = the AdjointMatrix divided by the MatrixDeterminant. Both the AdjointMatrix and the MatrixDeterminant are calculated recursively. For 2 x 2 matrices, these are easy to calculate. As the matrices get larger, these calculations become very tedious very quickly.

For a 2 x 2 matrix:

So, for the example on the MatrixFactoring page, we get:

The general case of finding a MatrixInverse can be solved by augmenting an IdentityMatrix? and using GaussianElimination

CategoryMath

The IdentityMatrix? is an n x n matrix with 1's along the main diagonal, and 0's everywhere else. When an n x n matrixB*B^(-1) =I

For a 2 x 2 matrix:

[ b11 b12 ]The adjoint matrix is:B= [ ] [ b21 b22 ]

[ b22 -b12 ] adj(The determinant is:B) = [ ] [-b21 b11 ]

det(B) = b11 * b22 - b12 * b21

So, for the example on the MatrixFactoring page, we get:

[ 0.795 8.805 ]B= [ ] [ 0.205 -7.805 ] [-7.805 -8.805 ] adj(B) = [ ] [-0.205 0.795 ] det(B) = (0.795)*(-7.805) - (0.205)*(8.805) = -8.01 [ 0.9744 1.09925]B^(-1) = [ ] [ 0.0256 -0.09925] [ 0.795 8.805 ] [ 0.9744 1.09925] [ 1 0 ] [ ] * [ ] = [ ] [ 0.205 -7.805 ] [ 0.0256 -0.09925] [ 0 1 ]

The general case of finding a MatrixInverse can be solved by augmenting an IdentityMatrix? and using GaussianElimination

CategoryMath

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