Cauchy expansion of determinant of a bordered matrix

[ekkilop]

The Cauchy expansion says that

\text{det} \begin{bmatrix}<br /> A &amp; x \\[0.3em]<br /> y^T &amp; a<br /> \end{bmatrix}<br /> = a \text{det}(A) - y^T \text{adj}(A) x,

where A is an n-1 by n-1 matrix, y and x are vectors with n-1 elements, and a is a scalar.
There is a proof in Matrix Analysis by Horn and Johnson that seems to be based on that A is a principal submatrix. My question is whether some similar result holds if A is not a principal submatrix? Say that we look for

det\begin{bmatrix}<br /> y^T &amp; a \\[0.3em]<br /> A &amp; x<br /> \end{bmatrix}<br />.

Would a similar expression hold?

Thanks.

 

[Mandelbroth]

The Cauchy expansion says that

\text{det} \begin{bmatrix}<br /> A &amp; x \\[0.3em]<br /> y^T &amp; a<br /> \end{bmatrix}<br /> = a \text{det}(A) - y^T \text{adj}(A) x,

where A is an n-1 by n-1 matrix, y and x are vectors with n-1 elements, and a is a scalar.
There is a proof in Matrix Analysis by Horn and Johnson that seems to be based on that A is a principal submatrix. My question is whether some similar result holds if A is not a principal submatrix? Say that we look for

det\begin{bmatrix}<br /> y^T &amp; a \\[0.3em]<br /> A &amp; x<br /> \end{bmatrix}<br />.

Would a similar expression hold?

Thanks.

Indeed. In fact, it would just be $\vec{y}^T \operatorname{adj}\textbf{A} \vec{x} - a\operatorname{det}\textbf{A}$. Can you see why?

 

[ekkilop]

Hi!

It just dawned on me that any such matrices (I suppose there are only 4 places A could go ^^, ) are related by simple permutations. Since any permutation matrix has determinant + or - 1 then what you say must be true.

Thank you for the enlightenment! =)