Lesson 4: Gaussian Reduction

New Concept

Gaussian reduction is a systematic method for solving a $3 \times 3$ linear system. It uses linear combinations to transform the system into a simpler triangular form, which can then be solved efficiently using back-substitution.

What’s next

Next, you'll master this method through interactive examples and practice problems, learning to solve for the unique intersection point of three planes.

Property

A solution to an equation in three variables, such as $x + 2y - 3z = -4$ is an ordered triple of numbers that satisfies the equation.

A solution to a system of three linear equations in three variables is an ordered triple that satisfies each equation in the system.

An ordered triple $(x, y, z)$ can be represented geometrically as a point in space using a three-dimensional Cartesian coordinate system, as shown in the figure.

Property

A special case is called back-substitution. It works when one of the equations involves exactly one variable, and a second equation involves that same variable and just one other variable. A $3 \times 3$ system with these properties is said to be in triangular form.

Examples

• Solve the system: $x+y+z=8$, $y-z=1$, and $2z=4$. From the third equation, $z=2$. Substitute into the second: $y-2=1$, so $y=3$. Substitute both into the first: $x+3+2=8$, so $x=3$. The solution is $(3, 3, 2)$.

• Use back-substitution on the system $2x-y+z=5$, $3y+z=11$, and $z=2$. Substitute $z=2$ into the second equation to find $3y+2=11$, so $y=3$. Substitute both values into the first: $2x-3+2=5$, so $x=3$. The solution is $(3, 3, 2)$.

Property

The method for solving a general $3 \times 3$ linear system is called Gaussian reduction. We use linear combinations to reduce the system to triangular form, and then use back-substitution to find the solutions.

Steps for Solving a $3 \times 3$ Linear System.

1. Clear each equation of fractions and put it in standard form.
2. Choose two of the equations and eliminate one of the variables by forming a linear combination.
3. Choose a different pair of equations and eliminate the same variable.
4. Form a $2 \times 2$ system with the equations found in steps (2) and (3). Eliminate one of the variables from this $2 \times 2$ system by using a linear combination.
5. Form a triangular system by choosing among the previous equations. Use back-substitution to solve the triangular system.

Examples

• To start solving $x+y+z=2$ and $2x-y+z=1$, eliminate $x$. Multiply the first equation by $-2$ to get $-2x-2y-2z=-4$. Add this to the second equation to get $-3y-z=-3$.

Property

If, at any step in forming linear combinations, we obtain an equation of the form

then the system is inconsistent and has no solution. If we don't obtain such an equation, but we do obtain one of the form

then the system is dependent and has infinitely many solutions.

Examples

• Consider the system $x+y+z=5$ and $x+y+z=1$. Subtracting the second from the first gives $(x+y+z)-(x+y+z) = 5-1$, which results in $0=4$. This contradiction means the system is inconsistent.

• Consider the system $2x-y=3$ and $4x-2y=6$. Multiply the first equation by $-2$ to get $-4x+2y=-6$. Adding this to the second equation gives $0=0$. This indicates the system is dependent.