Answer to Question #311847 in Discrete Mathematics for Showi

Question

Solve for the determinant in the equation below

2 −2 1

2 2 1

4 1 3

Answer 1

Solution

$det(A)=\begin{vmatrix} 2 & -2 & 1\\ 2 & 2 & 1\\ 4 & 1 & 3\\ \end{vmatrix}$

$det(A)=2\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}-(-2)\begin{vmatrix} 2 &1 \\ 4 & 3 \end{vmatrix}+(1)\begin{vmatrix} 2 & 2 \\ 4 & 1 \end{vmatrix}$

$det(A)=(2)(6-1)+(2)(6-4)+(1)(2-8)$

$det(A)=(2)(5)+(2)(2)+(1)(-6)$

$det(A)=10+4-6$

$det(A)=8$

Hence

$\begin{vmatrix} 2 & -2 & 1\\ 2 & 2 & 1\\ 4 & 1 & 3\\ \end{vmatrix}=8$