What is the determinant of a 3x3 matrix using various methods?

[karush]

$\textsf{Compute the determinant of} $
$$A=\left|
\begin{array}{rrr} 1&0&2\\ 1&0&0 \\ 3&2&0\end{array}
\right|$$
$\textsf{(a)by method of Basket weaving}$
$\begin{array}{rrrrr}
1&0&2&1&0 \\ 1&0&0&1&0\\ 3&2&0&3&2
\end{array}$
$[(1)(0)(0)+(0)(0)(3)+(2)(1)(2)]-[(3)(0)(2)+(2)(0)(1)+(0)(1)(0)]$
$[4]-[0]=4$
$\textit{(b) by cofactor expression }$
$\quad\textit{$C_3$ has 2 zeros so}$
$2\begin{vmatrix}1&0\\3&2\end{vmatrix}=4$ hopefully ok
suggestions

 

[tkhunny]

$\textsf{Compute the determinant of} $
$$A=\left|
\begin{array}{rrr} 1&0&2\\ 1&0&0 \\ 3&2&0\end{array}
\right|$$
$\textsf{(a)by method of Basket weaving}$
$\begin{array}{rrrrr}
1&0&2&1&0 \\ 1&0&0&1&0\\ 3&2&0&3&2
\end{array}$
$[(1)(0)(0)+(0)(0)(3)+(2)(1)(2)]-[(3)(0)(2)+(2)(0)(1)+(0)(1)(0)]$
$[4]-[0]=4$
$\textit{(b) by cofactor expression }$
$\quad\textit{$C_3$ has 2 zeros so}$
$2\begin{vmatrix}1&0\\3&2\end{vmatrix}=4$ hopefully ok
suggestions

Do C2 or R2, just for practice.

 

[HOI]

Yet another way, using "row reduction": From \(\displaystyle \begin{bmatrix}1 & 0 & 2 \\ 1 & 0 & 0 \\ 3 & 2 & 0 \end{bmatrix}\) subtract the first row from the second row and three times the first row from the third row to get \(\displaystyle \begin{bmatrix}1 & 0 & 2 \\ 0 & 0 & -2 \\ 0 & 2 & -6\end{bmatrix}\). Then swap the second and third rows to get the upper triangular matrix \(\displaystyle \begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & -6 \\ 0 & 0 & -2\end{bmatrix} which obviously has determinant \(\displaystyle (1)(2)(-2)= -4\). Adding or subtracting one row from another does not change the determinant while swapping two rows multiplies the determinant by -1. Since we swapped rows once, the determinant of the original matrix is 4.

(The other "row operation", multiplying a row by a number, multiplies the determinant by that number.)\)