The Determinant of a 3×3 Matrix

Find $\det \left(\begin{array}{ccc}2& -1& 3\\ -2& 7& 0\\ 4& 3& 4\end{array}\right)$.

1. First term (+): $2\times \det \left(\begin{array}{cc}7& 0\\ 3& 4\end{array}\right)=2(28-0)=56$ 2. Second term (−): $-(-1)\times \det \left(\begin{array}{cc}-2& 0\\ 4& 4\end{array}\right)=+1\times (-8-0)=-8$ Careful: the formula already has a minus before $b$, and here $b=-1$, so $-b=-(-1)=+1$. 3. Third term (+): $3\times \det \left(\begin{array}{cc}-2& 7\\ 4& 3\end{array}\right)=3(-6-28)=3(-34)=-102$ 4. Total: $56+(-8)+(-102)=-54$

Answer: $-54$

Find $\det \left(\begin{array}{cc}4& 4\\ 3& 6\end{array}\right)$ first as a warm-up: $24-12=12$.

Now $\det \left(\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 0& 6\end{array}\right)$:

• 1$+1\times \det \left(\begin{array}{cc}4& 5\\ 0& 6\end{array}\right)=24-0=24$
• 2$-2\times \det \left(\begin{array}{cc}0& 5\\ 1& 6\end{array}\right)=-2(0-5)=-2(-5)=10$
• 3$+3\times \det \left(\begin{array}{cc}0& 4\\ 1& 0\end{array}\right)=3(0-4)=-12$
• 4Total: $24+10-12=22$

Answer: $22$