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The inverse of a square matrix M is a matrix, denoted as M-1, with the property that M-1 M = M M-1 = I. Here I is the identity matrix of the same size as M, having 1's on the diagonal and 0's elsewhere.
In terms of transformations, M-1 undoes the transformation produced by M and so the combination M-1M represents the transformation that changes nothing.
The condition MM-1 = I can be written as
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and
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when k and i are different, and these conditions completely determine the matrix M-1 given M, when M has an inverse.
These equations have the same form as the two conditions (A) and (B) of section 4.3 except that det M is on the left-hand side in (A) instead of 1, and (-1)i + jMij appears in (A) and (B) instead of M-1ji here.
We can therefore divide both sides of (A) and (B) by det M, and deduce
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Remember that here Mij is the determinant of the matrix obtained by omitting the i-th row and j-th column of M; the elements of M are the mij, while M-1ji here represents the element of the inverse matrix to M in j-th row and i-th column.
We can phrase this in words as: the inverse of a matrix M is the matrix of its cofactors, with rows and columns interchanged, divided by its determinant.
Exercises:
4.7 Compute the inverse of the matrix in Exercise 4.4 using this formula. Check the product M-1M to be sure your result is correct.
4.8 Set up a spreadsheet that computes the inverse of any three by three
matrix with non-zero determinant, using this formula.
(Hint: by copying the first two rows into a fourth and fifth row and the first
two columns into a fourth and fifth column, you can make one entry and copy
to get all of the (-1)i + jMij at once. Then all that
is left is rearranging to swap indices and dividing by the determinant (which
is the dot product of any row of M with the corresponding cofactors).)
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