Find the determinant of the matrix $A=\left(\begin{array}{ccc}1& 2& 0\\ -3& 4& 5\\ 2& -1& 3\end{array}\right)$.

Step-by-step solution

1. Write the determinant using the first row expansion: $\det A=1\cdot \left|\begin{array}{cc}4& 5\\ -1& 3\end{array}\right|-2\cdot \left|\begin{array}{cc}-3& 5\\ 2& 3\end{array}\right|+0\cdot \left|\begin{array}{cc}-3& 4\\ 2& -1\end{array}\right|$.
2. Compute the 2$\times 2$ determinants: $\left|\begin{array}{cc}4& 5\\ -1& 3\end{array}\right|=4\cdot 3-5(-1)=12+5=17$ and $\left|\begin{array}{cc}-3& 5\\ 2& 3\end{array}\right|=(-3)\cdot 3-5\cdot 2=-9-10=-19$.
3. Substitute back: $\det A=1\cdot 17-2\cdot (-19)+0=17+38=55$.
4. Therefore the determinant equals $55$.

Concept tested: Determinant of a 3×3 matrix

Published 19 September 2026 · How this solution was prepared and checked