Evaluating a 3 by 3 Determinant Find the determinant of [latex]A=\left[\begin{array}{rrr}2 & 3 & 4 \\1 & -2 & 2 \\3 & 4 & -1\end{array}\right][/latex].

Expanding by the first row, we have [latex] |A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13},[/latex] where [latex]\begin{aligned}&A_{11}=(-1)^{1+1} M_{11}=(1)\left|\begin{array}{rr}-2 & 2 \\4 & -1\end{array}\right| \\&A_{12}=(-1)^{1+2} M_{12}=(-1)\left|\begin{array}{rr}1 & 2 \\3 & -1\end{array}\right| \\&A_{13}=(-1)^{1+3} M_{13}=(1)\left|\begin{array}{rr}1 & -2 \\3 & 4\end{array}\right|\end{aligned}[/latex] Thus, [latex]\left|\begin{array}{rrr}2 & 3 & 4 \\1 & -2 & 2 \\3 & 4 & -1\end{array}\right|=2(1)\left|\begin{array}{rr}-2 & 2 \\4 & -1\end{array}\right|+3(-1)\left|\begin{array}{rr}1 & 2 \\3 & -1\end{array}\right|+4(1)\left|\begin{array}{rr}1 & -2 \\3 & 4\end{array}\right|[/latex] [latex]=2(2-8)-3(-1-6)+4(4+6)[/latex] Evaluate determinants of order 2. = -12 + 21 + 40 = 49 Simplify.

Question 10.4.3: Evaluating a 3 by 3 Determinant Find the determinant of A = ...

Precalculus A Right Triangle Approach

Evaluating a 3 by 3 Determinant

Find the determinant of $A=\left[\begin{array}{rrr}2 & 3 & 4 \\1 & -2 & 2 \\3 & 4 & -1\end{array}\right]$ .

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Expanding by the first row, we have $|A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13},$ where

$\begin{aligned}&A_{11}=(-1)^{1+1} M_{11}=(1)\left|\begin{array}{rr}-2 & 2 \\4 & -1\end{array}\right| \\&A_{12}=(-1)^{1+2} M_{12}=(-1)\left|\begin{array}{rr}1 & 2 \\3 & -1\end{array}\right| \\&A_{13}=(-1)^{1+3} M_{13}=(1)\left|\begin{array}{rr}1 & -2 \\3 & 4\end{array}\right|\end{aligned}$

Thus,

$\left|\begin{array}{rrr}2 & 3 & 4 \\1 & -2 & 2 \\3 & 4 & -1\end{array}\right|=2(1)\left|\begin{array}{rr}-2 & 2 \\4 & -1\end{array}\right|+3(-1)\left|\begin{array}{rr}1 & 2 \\3 & -1\end{array}\right|+4(1)\left|\begin{array}{rr}1 & -2 \\3 & 4\end{array}\right|$

$=2(2-8)-3(-1-6)+4(4+6)$ Evaluate determinants of order 2.

= -12 + 21 + 40 = 49 Simplify.