Question #1ac69

This set of linear equations could be solved by various methods including 1. equation manipulation and variable elimination, 2. matrix inversion, 3. augmented matrix, etc. Let's try 1:

#color(red)(-6r+5s+2t=-11) color(white)"&&&&&&&&&&&&"(1)#
#color(blue)(-2r+s+4t=-9) color(white)"&&&&&&& & & & &&"(2)#
#color(magenta)(4r-5s+5t=-4) color(white)"&&&&&&& & & &&&&"(3)#

To begin, we'll eliminate #r# using two different pair of equations. Take #(1) - 3*(2)#:

#color(red)(-6r) color(blue)(+ 6r) color(red)(+5s) color(blue)(-3s) color(red)(+ 2t) color(blue)( -12t) = color(red)(-11) color(blue)( + 27)#

#2s - 10t = 16 color(white)"&&&&&&&&&&&&&&&&&"(4)#

Take #2*(2)+(3)#:

#color(blue)(-4r) color(magenta)(+4r) color(blue)(+2s) color(magenta)(-5s) color(blue)(+ 8t) color(magenta)(+5t) = color(blue)(-18) color(magenta)( -4)#

#-3s + 13t = -22 color(white)"&&&&&&&&&&&&&&&"(5)#

Now eliminate #s# by taking #3*(4)+2*(5)#:

#6s-6s-30t+26t=48-44#

#-4t=4#

#t=-1#

Let's substitute #t# back into #(1)# and #(2)#:

#-6r+5s-2=-11#
#-6r+5s = -9 color(white)"&&&&&&&&&&&&&&&"(6)#

#-2r+s-4=-9#
#-2r+s=-5 color(white)"&&&&&&&&&&&&&&&&"(7)#

Now take #(6) - 3*(7)#:

#-6r+6r+5s-3s = -9+15#

#2s=6#

#s=3#

Finally substitute #s# back into #(6)#

#-6r+15=-9#
#-6r=-24#
#r=4#

Summarizing:

#r=4; s=3; t=-1#