How do you solve 3y-z=-13y−z=−1, x+5y-z=-4x+5y−z=−4, -3x+6y+2z=11−3x+6y+2z=11 using matrices?

1 Answer
Alan P.
Feb 21, 2016

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

Explanation:

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

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

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

enter image source here

An interesting point to observe, because of the way computers perform (what is called "floating point") arithmetic the value of D_yDy shows up as being
color(white)("XXX")3E-16 = 0.0000000000000003XXX3E−16=0.0000000000000003 instead of 00
which, in turn, cause yy to display as
color(white)("XXX")-1.9E-17=-0.000000000000000019XXX−1.9E−17=−0.000000000000000019 instead of 00

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

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