How do you find the determinant of #((5, 2, 4), (2, -2, -3), (-3, 2, 5))#?

1 Answer
Jun 3, 2016

Answer:

It is #-30#.

Explanation:

For a #2\times2# matrix the determinant is

#det[((a,b),(c,d))]=ad-bc#

In a #3\times3# matrix you have to calculate the determinant of three sub-matrices #2\times2#.

We begin selecting the first element

#((\color[red]a,b,c),(d,e,f),(g,h,i))#

then we identify the sub-matrix that is not in the raw or the column of that element

#((\color[red]a,b,c),(d,\color[cyan]e,\color[cyan]f),(g,\color[cyan]h,\color[cyan]i))#

and finally we multiply the element selected (in our case #a#) by the determinant of the #2\times2# matrix

#a(ei-fh)#.

We repeat this process for the elements of the first raw alternating the signs

#((a,\color[red]b,c),(\color[cyan]d,e,\color[cyan]f),(\color[cyan]g,h,\color[cyan]i))#

#-b(di-fg)#

#((a,b,\color[red]c),(\color[cyan]d,\color[cyan]e,f),(\color[cyan]g,\color[cyan]h,i))#

#c(dh-g e)#.

The determinant is then the sum of these three terms:

#a(ei-fh)-b(di-fg)+c(dh-g e)#

We are now ready to calculate our determinant:

#det((5,2,4),(2,-2,-3),(-3,2,5))=#

#5*(-2*5-(-3)*2)#

#-2*(2*5-(-3)*(-3))#

#+4*(2*2-(-2)*(-3))#

#=-20-2-8=-30#.