Let A=pmatrixm&n&qpmatrix, d=|A|0, and A-d(adjA)=O.

Question

Let $A=\begin{pmatrix}m&n\p&q\end{pmatrix}$, $d=|A|
e0$, and $A-d(\operatorname{adj}A)=O$. Then:

Accepted Answer

$A-d(\operatorname{adj}A)=\begin{pmatrix}m-qd & n(1+d)\ p(1+d) & q-md\end{pmatrix}=O$ means, since $\det$ of this matrix is zero identically as $A$ satisfies this:

$(m-qd)(q-md)-np(1+d)^2=0 \Rightarrow mq-m^2d-q^2d+mqd^2-np(1+d)^2=0.$

Using $mq-np=d$: $d+d^2\cdot d - d(m^2+q^2) - np\cdot2d = 0$ leads (after simplification, as in the source) to
$$1+d^2=(m+q)^2-2d \Rightarrow (1+d)^2=(m+q)^2.$$

Solutions · 1