How do you solve #((-3, -2), (1, 1))X=((-8, -1),(6,0))#?

1 Answer
May 30, 2016

Answer:

#color(green)(=>X = ((-4,+1),(+10,-1)))#

Explanation:

Deliberately not using the shortcut method of using the determinant.

Write as # AX=B#

Then #A^(-1) AX=A^(-1)B#

But # A^(-1)A=I#

#=> X=A^(-1)B#

# color(red)("Note that the order of the matrices is important")#
#color(magenta)(A^(-1) A!=A A^(-1))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "A^(-1))#

#((-3,-2" |" 1,0),(" "1, 1 color(white)(..)" |"0,1))#
'...............................................

#Row( 1) + 3Row(2)#

#((0,1" |" 1,3),(1, 1 " |"0,1))#
'....................................................

#Row(2)-Row(1)#

#((0,color(white)(...)1" |"color(white)(..) 1,3),(1,color(white)(..) 0 " |"-1,-2))#
'................................................

Reverse the order of the rows

#((1,color(white)(..) 0 " |"-1,-2),(0,color(white)(.)1" |"color(white)(..) 1,3))#
'.......................................................

#color(brown)(=>A^(-1)=((-1,-2),(1,3)))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#=> X = A^(-1)B#

#=> X = ((-1,-2),(1,3))((-8,-1),(6,0))#

#"Let " X = ((a,b),(c,d))#

#a=[(-1)xx(-8)]" "+" "[(-2)xx(6) ]= -4#

#b=[(-1)xx(-1)]" "+" "[(-2)xx(0)] = +1#

#c=[(1)xx(-8)]" "+" "[(3)xx(6)] = +10#

#d=[(1)xx(-1)]" "+" "[(3)xx(0)]=-1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(green)(=>X = ((-4,+1),(+10,-1)))#