# How do you find the determinant of ((-5, 4, 1), (4, 7, 0), (-3, 4, -1))?

Jun 13, 2016

The determinant of the given matrix is $88.$

#### Explanation:

If $A$ is a matrix, where $A = \left[\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}\right]$, then we define the determinant of the matrix to be

$| A | = | \left({a}_{11} , {a}_{12} , {a}_{13}\right) , \left({a}_{21} , {a}_{22} , {a}_{23}\right) , \left({a}_{31} , {a}_{32} , {a}_{33}\right) |$

$\implies | A | = {a}_{11} | \left({a}_{22} , {a}_{23}\right) , \left({a}_{32} , {a}_{33}\right) | - {a}_{12} | \left({a}_{21} , {a}_{23}\right) , \left({a}_{31} , {a}_{33}\right) | + {a}_{13} | \left({a}_{21} , {a}_{22}\right) , \left({a}_{31} , {a}_{32}\right) |$

$\implies | A | = {a}_{11} \left({a}_{22} {a}_{33} - {a}_{32} {a}_{23}\right) - {a}_{12} \left({a}_{21} {a}_{33} - {a}_{31} {a}_{23}\right) + {a}_{13} \left({a}_{21} {a}_{32} - {a}_{31} {a}_{22}\right)$

And then simply this expression.

Now, let $A = \left[\begin{matrix}- 5 & 4 & 1 \\ 4 & 7 & 0 \\ - 3 & 4 & - 1\end{matrix}\right]$

$\implies | A | = | \left(- 5 , 4 , 1\right) , \left(4 , 7 , 0\right) , \left(- 3 , 4 , - 1\right) | = - 5 | \left(7 , 0\right) , \left(4 , - 1\right) | - 4 | \left(4 , 0\right) , \left(- 3 , - 1\right) | + 1 | \left(4 , 7\right) , \left(- 3 , 4\right) |$

$\implies | A | = - 5 \left\{7 \cdot \left(- 1\right) - 4 \cdot 0\right\} - 4 \left\{4 \left(- 1\right) - \left(- 3\right) \cdot 0\right\} + 1 \left\{4 \cdot 4 - \left(- 3\right) \cdot 7\right\}$

$\implies | A | = - 5 \left(- 7 - 0\right) - 4 \left(- 4 - 0\right) + \left(16 + 21\right)$

$\implies | A | = 35 + 16 + 37 = 88$

Hence, determinant of the given matrix is $88.$