
Kenneth S. answered 02/21/16
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Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
Here's how to do problem 2 using Cramer's rule.
Create matrix B on the calculator, containing 3 columns (coefficients of x, y, x) and 3 rows (three equations):
1 -2 -5
-2 6 18
-3 8 20
then det([B]) = -6 <--this is the value of the System Determinant, D.
Now edit [B] to put the right hand side constants only in place of the x-column coefficients:
4 -2 -5
-8 6 18
-18 8 20
then det([B]) =12. <--this is the value of the determinant Dx
-8 6 18
-18 8 20
then det([B]) =12. <--this is the value of the determinant Dx
So the value of x = 12/-6 = -2. [x = Dx / D].
Now edit [b] to contain these data,:
1 - 4- 5
-2 -8 18
-3 -18 20
then det([B]) = 48 <-- Dy. The value of y = Dy / D so y =48 / -6 = -8.
-2 -8 18
-3 -18 20
then det([B]) = 48 <-- Dy. The value of y = Dy / D so y =48 / -6 = -8.
Finally, create the matrix [B] in which the right-side constants are substituted for the column of "z" coefficients:
1 -2 4
-2 6 -8
-3 8 -18
then det([B]) = -12 <--Dz. So z = Dz / D = -12 / -6 = 2.
-2 6 -8
-3 8 -18
then det([B]) = -12 <--Dz. So z = Dz / D = -12 / -6 = 2.
Solution: (-2, -8, 2)
[verified using rref command on 3 by 4 Augmented Matrix for this system]