How do you find the determinant of #((-1, -3, -3), (4, 4, 1), (-2, 0, 3))#?

Question

1243 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-18T09:56:27

#D=6#

#((-1,-3,-3),(4,4,1),(-2,0,3))=#
#(-1)*((4,1),(0,3))-(-3)((4,1),(-2,3))+(-3)((4,4),(-2,0))#

#=(-1)(4*3-0*1)+3(4*3-(-2)1)-3*(4*0-(-2)*4)#

#=(-1)(12-0)+3(12+2)-3*(0+8)#

#=-12+42-24#

#=6#

or by the formula for

#((a_11, b_12, c_13),(a_21, b_22, c_23),(a_31,b_32, c_33))#

#=a_11 b_22 c_3+b_12 c_23 a_31+c_13 b_32 a_21-a_31 b_22 c_13-b_32 c_23 a_11-c_33 b_12 a_21#

so that

#((-1,-3,-3),(4,4,1),(-2,0,3))=#

#[(-1)( 4)(3)]+[(-3)( 1)( -2)]+ [(-3)( 0)(4)] -[( -2)(4 )(-3)] - [(0)(1)( -1)] -[( 3)( -3)(4)]#

#-12+6-0-24-0+36#

#=6#

have a nice day ! from the Philippines .