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Question 1

Lets consider a $3 \times 3$ matrix, A.

The inverse of this matrix can be found by finding the adjoint of the matrix and dividing it by its determinant. Can someone please explain why this process work and how it can be visualized?

I know that the determinant can be thought of as the volume spanned by the column vectors and that the adjoint of matrix A is $A^T$'s cofactors.

Question 2

Let's consider a $3\times 3$ matrix, A.

The cofactors of this matrix can be found by the cross product of each of its column vectors. So what does the cofactor matrix tell us?

Thank you.

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I think you mean the adjugate (sometimes called the "classical adjoint") of the matrix, not the adjoint. The adjoint is the conjugate transpose; the adjugate is the matrix that is obtained by replacing each entry of $A$ with its corresponding cofactor, and then transposing. And I'm not sure what you mean by "cross product of its column vectors. The cross product needs two arguments, so "each" makes no sense, and it gives a vector, so certainly not equal to the cofactor. What do you mean? –  Arturo Magidin Nov 27 '10 at 23:23
    
@Arturo She does seem to mean adjugate. As for the other point, it's ok. Take 2 vectors out of 3, get the cross product vector. In total, 3 vectors - the columns of the adjugate. –  Max Nov 28 '10 at 0:15
    
@Arturo: Well, it depends how old you are. What you call the adjoint is the Hermitian adjoint, and what you call the adjugate is the "classical" adjoint. Many older texts use adjoint in the sense of the question. –  Geoff Robinson Jul 8 '11 at 21:55

1 Answer 1

up vote 1 down vote accepted

There is some explanation in the Geometric Interpretation here, but I don't really like it too much.

I think it's easier to think this way: The cross product of $v$ and $w$ is a vector $x$ orthogonal to the two given ones, whose length is twice the area of the triangle. Hence the volume of the tetrahedron on $u, v, w$ is one sixth of the the dot product of $x$ with the remaining vector $u$ (one should be careful with signs, but it works out). The dot product of $x$ with $v$ or $w$ is zero. Similar things are true about the cross product of $u$ and $w$ dotted with $v$ (and $u$ and $w$) and the cross product of $u$ and $v$ dotted with $w$ (and $u$ and $v$). What all of this implies is that the matrix of cross product vectors (transposed) multiplied with the matrix of $u$, $v$, $w$ is diagonal, with all diagonal entries equal to six times the volume of the tetrahedron, i.e. the determinant of the $u$, $v$, $w$ matrix. The result follows.

Higher dimensional versions are fairly similar, with volumes of $n$ and $n-1$ simplices.

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An algebraic way to see this (which works for larger matrices) is to use Cramer's rule, and use the fact that a matrix with repeated rows has zero determinant. Cramer's rule is computationally cumbersome, but theoretically useful. –  Geoff Robinson Jul 8 '11 at 22:01

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