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

Step-by-step solution

1. Write the matrix $B$ and recall that the determinant of a $3\times 3$ matrix can be computed by expanding along the first row.
2. Compute $\det (B)=1\cdot \det \left(\begin{array}{cc}-1& 4\\ 0& 2\end{array}\right)-2\cdot \det \left(\begin{array}{cc}0& 4\\ 5& 2\end{array}\right)+3\cdot \det \left(\begin{array}{cc}0& -1\\ 5& 0\end{array}\right)$.
3. Evaluate each $2\times 2$ determinant: $\det \left(\begin{array}{cc}-1& 4\\ 0& 2\end{array}\right)=(-1)(2)-0\cdot 4=-2$, $\det \left(\begin{array}{cc}0& 4\\ 5& 2\end{array}\right)=0\cdot 2-5\cdot 4=-20$, $\det \left(\begin{array}{cc}0& -1\\ 5& 0\end{array}\right)=0\cdot 0-5\cdot (-1)=5$.
4. Substitute: $\det (B)=1(-2)-2(-20)+3(5)=-2+40+15=53$.
5. Since a sign error was made in the expansion, recompute carefully: the correct expansion gives $\det (B)=1(-2)-2(-20)+3(5)=-2+40+15=53$.
6. Thus the determinant is $53$.
7. (If a different sign convention was used, the final value is $53$).

Concept tested: Determinant of a 3x3 matrix

Published 6 September 2026 · How this solution was prepared and checked