|1 o |
| 0 2 |
= (2x1)-(0x0)=2
|1 -2|
|-1 3|
= (1x3)-(-2x-1)=1
| 2 -3 -2|
| -1 3 4 |
| -7 -2 8 |
2| 3 4 | X 3 | -1 4 | x -2 | -1 3|
|-2 8| | -7 8 | |-7 -2|
= 2(8-24)+3(8+28)-2(2+21)
= 94
did i do these right ?
|1 o |
| 0 2 |
= (2x1)-(0x0)=2
|1 -2|
|-1 3|
= (1x3)-(-2x-1)=1
| 2 -3 -2|
| -1 3 4 |
| -7 -2 8 |
2| 3 4 | X 3 | -1 4 | x -2 | -1 3|
|-2 8| | -7 8 | |-7 -2|
= 2(8-24)+3(8+28)-2(2+21)
= 94
did i do these right ?
1 and 2 are correct. You need to remember that the signs alternate when evaluating a 3x3 determinant or higher. A diagram like this helps:
$\displaystyle \displaystyle \begin{align*} \left| \begin{matrix} + & - & + & - & \dots \\ - & + & - & + & \dots \\ + & - & + & - & \dots \\ - & + & - & + & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} \right| \end{align*}$
So to evaluate your 3x3 determinant, it would be
$\displaystyle \displaystyle \begin{align*} \left| \begin{matrix} \phantom{-}2 & -3 & -2 \\ -1 & \phantom{-}3 & \phantom{-}4 \\ -7 & -2 & \phantom{-}8 \end{matrix} \right| &= 2\left| \begin{matrix} \phantom{-}3 & 4 \\ -2 & 8 \end{matrix} \right| - (-3) \left| \begin{matrix} -1 & 4 \\ -7 & 8 \end{matrix} \right| + (-2) \left| \begin{matrix} -1 & \phantom{-}3 \\ -7 & -2 \end{matrix} \right| \end{align*}$
there are 2 ways to evaluate a 3x3 determinant:
way #1) expansion by minors:
$\displaystyle \det(A) = a_{11}\det(A_{1,1}) - a_{12}\det(A_{1,2}) + a_{13}\det(A_{1,3})$
$\displaystyle = a_{11}\det(A_{1,1}) - a_{21}\det(A_{2,1}) + a_{31}\det(A_{3,1})$
where $\displaystyle A_{i,j}$ is the 2x2 matrix obtained by removing row i and column j.
way #2) the rule of sarrus:
$\displaystyle \det(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33} - a_{11}a_{23}a_{32}$
(often displayed as three diagonal products each way that "wrap around")
let us calculate:
$\displaystyle \det(A) = \begin{vmatrix}1&3&-2\\5&1&4\\0&2&1 \end{vmatrix}$
using both methods.
the first way, we have a choice: expansion by a row, or by a column (we could use ANY row or column, but we'll use the first one to keep it simple). first we'll expand by the first row:
to do this we need to find the 2x2 determinants, first.
$\displaystyle \det(A_{1,1}) = \begin{vmatrix} 1&4\\2&1 \end{vmatrix} = 1 - 8 = -7$
$\displaystyle \det(A_{1,2}) = \begin{vmatrix} 5&4\\0&1 \end{vmatrix} = 5 - 0 = 5$
$\displaystyle \det(A_{1,3}) = \begin{vmatrix} 5&1\\0&2 \end{vmatrix} = 10 - 0 = 10$
thus $\displaystyle \det(A) = (1)(-7) - (3)(5) + (-2)(10) = -7 - 15 - 20 = -42$.
now we'll expand by the first column. we need to calculate 2 more 2x2 determinants (we'll re-use the one obtained by eliminating row 1 and column one):
$\displaystyle \det(A_{2,1}) = \begin{vmatrix} 3&-2\\2&1 \end{vmatrix} = 3 - (-4) = 7$
$\displaystyle \det(A_{3,1}) = \begin{vmatrix} 3&-2\\1&4 \end{vmatrix} = 12 -(-2) = 14$
thus $\displaystyle \det(A) = (1)(-7) - 5(7) + 0(14) = -7 - 35 + 0 = -42$.
finally, we'll use the rule of sarrus:
$\displaystyle \det(A)$
$\displaystyle = (1)(1)(1) + (3)(4)(0) + (-2)(5)(2) - (-2)(1)(0) - (3)(5)(1) - (1)(4)(2)$
$\displaystyle = 1 + 0 - 20 - 0 - 15 - 8 = 1 - 43 = -42$.
personally, i like the rule of sarrus, but expansion by minors generalizes better to larger matrices.