1

I am attempting to complete the exercises in my textbook on matrix differential calculus. This question is giving me some trouble. These are the problems I am attempting to complete:

If $|A| \ne 0$, prove that

$$\begin{vmatrix} A & b \\ a' & \alpha\end{vmatrix}=(\alpha - a'A^{-1}b)|A|.$$

If $\alpha \ne 0$, prove that

$$\begin{vmatrix} A & b \\ a' & \alpha\end{vmatrix}=\alpha |A-\left( \frac1\alpha\right)ba'|$$

I found on wikipedia Determinant of Block Matrix which shows how if you have a partitioned matrix you can decompose that matrix into an upper and lower triangular matrix and apply the product rule to the determinant to find it. Seen as follows:

When $A$ is invertible, one has $$\det\begin{pmatrix} A & B \\ C & D \end{pmatrix}=\det(A) \times \det(D-CA^{-1}B)$$ as can be seen by employing the decomposition

$$\begin{pmatrix} A & B \\ C & D\end{pmatrix}=\begin{pmatrix} A & 0 \\ C & I_m\end{pmatrix}\begin{pmatrix} I_n & A^{-1}B \\ 0 & D-CA^{-1}B\end{pmatrix}$$

My question is how to extend this logic to a partitioned matrix where on wikipedia's example, $B$ and $C$ are vectors. Down the page a bit they state

When $D$ is a $1 \times 1$ matrix, $B$ is a column vector and $C$ is a row vector then

$$\det\begin{pmatrix} A & B \\ C & D \end{pmatrix}=\det(D-CA^{-1}B) \det(A)$$

which is essentially what I am trying to prove. I am just not quite sure how to prove this. I have attempted to use laplace expansion, but that didn't really get me anywhere. Not sure if this is because I am missing something, or if laplace expansion is not the proper technique to prove this.

Rodrigo de Azevedo
  • 23,223
Chives
  • 35

1 Answers1

1

Row and column matrices are special case of general matrices, the same rules apply.

Suppose you have accepted the fact that

$$\det\begin{pmatrix} A & B \\ C & D\end{pmatrix}=\det(A) \det(D-CA^{-1}B)$$

Then by letting $B=b$, $C=a'$ and $D=\alpha$,

We have if $\det(A) \ne 0$,

$$\det\begin{pmatrix} A & b \\ a' & \alpha\end{pmatrix}=\det(A) \det(\alpha-a'A^{-1}b)=(\alpha - a'A^{-1}b) |A|$$

If $\alpha \ne 0$,

\begin{align}\det\begin{pmatrix} A & b \\ a' & \alpha\end{pmatrix}&=\det \left( \begin{pmatrix} 0^T & 1 \\ I_{n-1} & 0\end{pmatrix}\begin{pmatrix} \alpha & a' \\ b & A\end{pmatrix} \begin{pmatrix} 0^T & 1 \\ I_{n-1} & 0\end{pmatrix}^T\right)\\&=\det \left( \begin{pmatrix} 0^T & 1 \\ I_{n-1} & 0\end{pmatrix}\right)^2 \det\left(\begin{pmatrix} \alpha & a' \\ b & A\end{pmatrix} \right)\\&=\det\left(\begin{pmatrix} \alpha & a' \\ b & A\end{pmatrix} \right)\\ &=\alpha \det(A-b\left(\frac1\alpha\right) a')\end{align}

Siong Thye Goh
  • 153,832
  • so $$\det(\alpha-a'A^{-1}b)$$ is the determinant of a constant, is the determinant of a constant just equal to a constant, couldn't find that in the properties of determinants in my textbook. – Chives Apr 15 '20 at 23:52
  • nevermind, figured out that the determinant of a 1x1 is just the number. – Chives Apr 16 '20 at 02:22
  • yup, you can view it as a diagonal matrix, hence it is equal to the scalar itself. – Siong Thye Goh Apr 16 '20 at 02:27