 ### Solving Systems of Equations Using Determinants© 2004, A2

If
 Ax +By = C Dx +Ey = F
then
 CE-BF x= AE-BD
and
 AF-CD y = AE-BD

Assume that every component of a linear system that can be a variable is a variable.

Solve the system.

The solution is the solution for EVERY linear system.

The solution above is that solution and the work to prove it is below.

But, it may be easier to just do the algebra each time rather than remember for certain the formula above. That is ... unless you know an easy way to remember the formula. Determinants and Cramer's Rule provide that easy way to remember the required computation.

The graphic at the left introduces math vocabulary, which need not even be known. All you really need to know is the replacement technique displayed in the graphic below.

The special features of a system may be seen in the array, of coefficients and constants. These are arranged in columns, a column for each variable and a column for the constants.

A matrix, symbols [ ], declares an array to be a whole thing to be considered as one thing and having special functions very often treating a row or a column at a time.

One thing that might be done with a matrix or array of coefficients and constants is:

Evaluate a determinant.

To evaluate a 2 column by 2 row determinant, 2 by 2 determinant, two multiplications and one subtraction are required. See the required computation displayed in the graphic at the left. See how it plays into solving the system in the graphic below. Though the technique is usually reserved for work with 3 or more unknowns rather than with the two unknowns as shown above it does provide an easy way to solve a linear system with two unknowns.

Problems involving linear system solution are found on other pages linked through the Algebra Page.

The Linear System Solver Page goes through a step-by-step solution using determinants.

Click on this problem, , to see the work and solution.

 Solve: Ax +By = C Dx +Ey = F

 -D(Ax +By = C) A(Dx +Ey = F)

 -ADx -BDy = -CD ADx +AEy = AF -ADx +ADx -BDy+AEy = -CD+AF (-BD+AE)y =(-CD+AF) (-BD+AE) (-BD+AE) (-CD+AF) y = (-BD+AE) simplified: AF-CD y = AE-BD

 Solve: Ax +By = C Dx +Ey = F

 E(Ax +By = C) -B(Dx +Ey = F)

 AEx +BEy = CE -BDx -BEy =-BF

 AEx +BEy = CE -BDx -BEy =-BF AEx -BDx +BEy-BEy = CE-BF (AE-BD)X =(CE-BF) (AE-BD) (AE-BD) (CE-BF) x= (AE-BD) simplified: CE-BF x= AE-BD                    © January 27, 2004www.mathnstuff.com/math/algebra/qb/adeterm.htm