How do you solve #3y-z=-1#, #x+5y-z=-4#, # -3x+6y+2z=11# using matrices?

1 Answer
Feb 21, 2016

Answer:

Use Cramer's Rule (with Determinants) to get
#color(white)("XXX")(x,y,z)=(-3,0,1)#

Explanation:

Using the coefficient of #x, y, and z# plus the equated constants as columns we can write these equations in matrix form.

If #D# is the determinant of the variable coefficient matrix
and #D_a, ain{x,y,z}# is the determinant of the variable coefficient matrix with the column for variable #a# replaced by the equated constants column,

Cramer's Rule tells us that:
#color(white)("XXX")a=D_a/D# for #ain{x,y,z}#

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An interesting point to observe, because of the way computers perform (what is called "floating point") arithmetic the value of #D_y# shows up as being
#color(white)("XXX")3E-16 = 0.0000000000000003# instead of #0#
which, in turn, cause #y# to display as
#color(white)("XXX")-1.9E-17=-0.000000000000000019# instead of #0#

You must be prepared to apply reasonable interpretations when working with computer generated outputs.