determinant with adjunct of matrix of n order

Question

Let be $A\in M_{n}(\mathbb{C})$, $L\in M_{1,n}(\mathbb{C})$, $C\in M_{n,1}(\mathbb{C})$.

Prove that $det(A-CL)=det(A)-LA^{*}C$, where $A^{*}$ is the adjunct of A.

Answer 1

You may start from $LA^*$. $A^*$ is the transpose of the cofactor matrix of $A$.

It is helpful to notice that if $A_i$ is the i-th row of $A$ and $A_{i}^*$ is the i-th column of $A^*$then we have

$A_iA_{i}^*=det(A)$

Now, we need to see how to use it to rewrite $LA^*$

$LA^*=[det(A(1,L)),det(A(2,L)),...,det(A(n,L))]$

where $A(j,L)$ is the matrix $A$, with its j-th row replaced by $L$.

Note that $LA^*$ is a row vector now.

Multiply it by $C$ gives

$-LA^*C=-c_1\times det(A(1,L)) -c_2\times det(A(2,L))-...-c_n\times det(A(n,L))]$

$-LA^*C=-\sum_{i} c_i\times det(A(i,L))$

$c_i \hspace{0.2cm} , (i=1,2,...,n)$ are elements of matrix $C$, in the same order as in the vector.

There is a property of the determinant, which says $c\times det(X)$ is equal to the determinant of the matrix that is obtained by multiplying one arbitrary row of $X$ by $c$, i.e. $c\times det(X)=det(X(j,cX_j))$

$j$ is arbitrary and $X_j$ is the j-th row of $X$. Therefore

$-LA^*C=-\sum_{i} c_i\times det(A(i,L))=-\sum_{i} det(A(i,c_iL))$

The RHS becomes

$$det(A)-LA^*C=det(A)-\sum_{i} det(A(i,c_iL))$$

For the LHS, you should recall another property of the determinant, which say, if the j-th row of a matrix $X$ can be written as $X_j=X_j^{1}+X_j^{2}$, then

$det(X)=det(X(j,X_j^{1}))+det(X(j,X_j^{2}))$. Therefore, if $A-CL=T$, we have

$det(A-CL)=det(T)=det(T(1,A_1))+det(T(1,-CL_1))$. In the second term of RHS is

$det(T(1,-CL_1))= det(\begin{bmatrix} -CL_1 \\ A_2-CL_2 \\ \vdots \\ A_n-CL_n \end{bmatrix})=det(\begin{bmatrix} -CL_1 \\ A_2 \\ \vdots \\ A_n \end{bmatrix})=-det(A(1,c_1L))$

Just because you can add any multiple of a row to other rows, without changing the determinant.

You should do the same process to the second row of the first term $det(T(1,A_1))$ and continue until you get $det(A)$ as one term.