Disclaimer: This answer might seem long, but I've shown nearly every step to the solution. When you get good at it, you can skip many of the steps shown
You can use echelon form where you list just the coefficients into an augmented matrix style format:
So:
#a_1x+b_1y+c_1z=d_1#
#a_2x+b_2y+c_2z=d_2#
#a_3x+b_3y+c_3z=d_3#
Would become in augmented matrix form:
#[(a_1, b_1, c_1, |d_1),(a_2, b_2, c_2, |d_2), (a_3, b_3, c_3, |d_3)]## {:("R1 (Row 1)"), ("R2 (Row 2)"), ("R3 (Row 3)"):}#
We need to get a triangle of #0#'s in the bottom left corner (echelon form) like :
#[(a_1, b_1, c_1, |d_1),(color(red)0, b_2, c_2, |d_2), (color(red)0, color(red)0, c_3, |d_3)] #
We do this by conducting Row Operations, which are pretty much addition, subtraction and scalar multiples of rows to get desired values. An example would be #"R1"-2"R2"->"R2"#, which means that the new row 2 is given by the previous row 1 minus 2 times the previous row 2.
If we are using the general #a, b, c#, it would look like:
#[(a_1, b_1, c_1, |" "d_1),((a_1)-2(a_2), (b_1)-2(b_2), (c_1)-2(c_2), |(d_1)-2(d_2)), (a_3, b_3, c_3, |" "d_3)]#
So for the system we have:
#4x-y+1/2z=-7#
#1/2x+4y-4z=-9#
#x+4y+4z=6#
In augmented matrix form:
#[(4, -1, 1/2, |-7),(1/2, 4, -4, |-9), (1, 4, 4, | " "6)] #
The steps to reduce to echelon form are shown below:
#=>[(4, -1, 1/2, |-7),(1/2, 4, -4, |-9), (1, 4, 4, | " "6)] # #{:(" "), ("R1"-8"R2"->"R2"), ("R1"-4"R3"->"R3"):}#
So you would do the following calculations:
For new row 2, you would do:
#a# value: #4-8(1/2)=0#
#b# value: #-1-8(4)=-33#
#c# value: #1/2-8(-4)=65/2=32.5#
#d# value: #-7-8(-9)=65#
For new row 3:
#a# value: #4-4(1)=0#
#b# value: #-1-4(4)=-17#
#c# value: #1/2-4(4)=-31/2=-15.5#
#d# value: #-7-4(6)=-31#
Our new matrix would be:
#=>[(4, -1, 1/2, |-7),(0, -33, 32.5, |" "65), (0, -17, -15.5, | -31)] #
Then do another row operation:
#=>[(4, -1, 1/2, |-7),(0, -33, 32.5, |" "65), (0, -17, -15.5, | -31)] # #{:(" "), (" "), (17"R2"-"33R3"->"R3"):}#
For new row 3:
#a# value: #17(0)-33(0)=0#
#b# value: #17(-33)-33(-17)=0#
#c# value: #17(32.5)-33(-15.5)=1064#
#d# value: #17(65)-33(-31)=2128#
So our new matrix would be:
#=>[(4, -1, 1/2, |-7),(0, -33, 32.5, |" "65), (0, 0, 1064, | 2128)] #
Now this is pretty much:
#4x-y+1/2z=-7# #" "#(Row 1)
#-33y+32.5z=65# #" "#(Row 2)
#1064z=2128##" "# (Row 3)
Using row 3:
#1064z=2128#
#=>z=2128/1064#
#=>z=2#
Now we know #z# and we can substitute it into row 2 and solve for #y#:
#=>-33y+32.5(2)=65#
#=>y=(65-32.5(2))/-33#
#=>y=0#
Now we can substitute both #y# and #z# into row 1 and solve for #x#:
#4x-(0)+1/2(2)=-7#
#=>x=(-2-2)/4#
#=>x=-2#
Hence you have the values for
#x=-2#
#y=0#
#z=2#
