Let $P=\left[\begin{array}{ccc}2& -1& 0\\ 1& 3& 4\\ 5& 2& -2\end{array}\right]$. Compute $det\left(\mathrm{adj}P\right)$.

Step-by-step solution

1. Recall the property: for an $n\times n$ matrix $A$, $det(\mathrm{adj}A)=\det (A{)}^{,n-1}$. Here $n=3$.
2. First find $detP$ by cofactor expansion along the first row.
3. Term1: $2\times \det \left[\begin{array}{cc}3& 4\\ 2& -2\end{array}\right]=2(3\cdot (-2)-4\cdot 2)=2(-6-8)=-28$.
4. Term2: $-(-1)\times \det \left[\begin{array}{cc}1& 4\\ 5& -2\end{array}\right]=+1(1\cdot (-2)-4\cdot 5)=-22$.
5. Term3 involves the zero entry and contributes $0$.
6. Add: $detP=-28-22=-50$.
7. Apply the property: $det(\mathrm{adj}P)=(-50{)}^{3-1}=(-50{)}^2=2500$.
8. Thus the required determinant is $2500$.

Concept tested: Determinant of an adjoint matrix

Published 17 September 2026 · How this solution was prepared and checked