How do you find the determinant of #((5, 7, -1, 1, 0, 0), (2, 2, 1, 3, -1, 0), (-3, 4, -1, -2, 1, 2), (2, 9, -3, 0, 0, 0), (0, 1, 4, 0, 0, 0), (0, 0, 1, 0, 0, 0))#?

1 Answer
Apr 28, 2016

Answer:

#4#

Explanation:

#[[color(teal)(5),color(teal)(7),color(teal)(-1),color(teal)(1),color(teal)(0),color(teal)(0)],[2,2,1,3,-1,0],[color(blue)(-3),color(blue)(4),color(blue)(-1),color(blue)(-2),color(blue)(1),color(blue)(2)],[2,9,-3,0,0,0],[0,1,4,0,0,0],[0,0,1,0,0,0]]#
#-># changing row1 with row3

#=-[[color(magenta)(-3),4,color(orange)(-1),-2,1,2],[color(magenta)(2),2,color(orange)(1),3,-1,0],[color(magenta)(5),7,color(orange)(-1),1,0,0],[color(magenta)(2),9,color(orange)(-3),0,0,0],[color(magenta)(0),1,color(orange)(4),0,0,0],[0,0,1,0,0,0]]#
#-># changing column1 with column3

#=[[-1,4,-3,-2,1,2],[1,2,2,3,-1,0],[-1,7,5,1,0,0],[-3,9,2,0,0,0],[4,1,0,0,0,0],[1,0,0,0,0,0]]#

#=-[1*1*2*1*(-1)*2]=4#

A brief explanation of the MINUS sign before the multiplication in the last line:
Considering the last determinant, how many times does one have to change columns to arrive at the shape

#[[a,b,c,d,e,f],[0,g,h,i,j,k],[0,0,l,m,n,o],[0,0,0,p,q,r],[0,0,0,0,s,t],[0,0,0,0,0,u]]#?

Answer: 3 times (column1 with column6, column2 with column5 and column3 with column4)
Each time the columns are changed with each other the sign is changed, so:

#(-1)^3=-1# (that's why the multiplication is preceded by a minus sign)