Equation with matrix and determinant

Question

Given

$$A={\begin{bmatrix} \det(A) & \det(A)+a \\ \det(A) +b & \det(A)+c \end{bmatrix}}$$

where $a,b,c$ are given and $A$ is unknown, is it possible to use some clever tricks concerning determinants for this case? (instead of direct calculations).

Answer 1

Let $\text{det}A=d$. Then we are given $$A={\begin{bmatrix} d & d+a \\ d +b & d+c \end{bmatrix}}$$ Using row operations, we get $$ d=|A|=\begin{vmatrix} d & d+a \\ d +b & d+c \end{vmatrix} \rightarrow \begin{vmatrix} d & d+a \\ b & c-a \end{vmatrix} \rightarrow \begin{vmatrix} d & a \\ b & c-a-b \end{vmatrix} $$ Thus $$d=d(c-a-b)-ab.$$ Now solve for $d$.

Answer 2

Let

$$\mathrm A = \det(\mathrm A) \, 1_2 1_2^{\top} + \begin{bmatrix} 0 & a\\ b & c\end{bmatrix}$$

Assuming that $a b \neq 0$ and using the matrix determinant lemma,

$$\begin{array}{rl} \det(\mathrm A) &= \left( 1 + \det(\mathrm A) 1_2^{\top} \begin{bmatrix} 0 & a\\ b & c\end{bmatrix}^{-1} 1_2 \right) \det \begin{bmatrix} 0 & a\\ b & c\end{bmatrix}\\ &= \left( 1 - \frac{\det(\mathrm A)}{a b} 1_2^{\top} \begin{bmatrix} c & -a\\ -b & 0\end{bmatrix} 1_2 \right) (- a b)\\ &= - a b + \det(\mathrm A) 1_2^{\top} \begin{bmatrix} c & -a\\ -b & 0\end{bmatrix} 1_2\\ &= (c - a - b) \det(\mathrm A) - a b\end{array}$$