1.10.2 Algebraic Solutions

Row Reduction

How can I write a system of linear equations?

To save space we can just write the coefficients without the variables
For 2 variables: $a_{1} x + b_{1} y = c_{1} a_{2} x + b_{2} y = c_{2}$ can be written shorthand as $[\begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{matrix} | \begin{matrix} c_{1} \\ c_{2} \end{matrix}]$
For 3 variables: $a_{1} x + b_{1} y + c_{1} z = d_{1} a_{2} x + b_{2} y + c_{2} z = d_{2} a_{3} x + b_{3} y + c_{3} z = d_{3}$ can be written shorthand as $[\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | \begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$

What is a row reduced system of linear equations?

A system of linear equations is in row reduced form if it is written as:

$[\begin{matrix} A_{1} & B_{1} & C_{1} \\ 0 & B_{2} & C_{2} \\ 0 & 0 & C_{3} \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ D_{3} \end{matrix}]$ which corresponds to $\begin{array}{rcl} A_{1} x + B_{1} y + C_{1} z & = & D_{1} \\ B_{2} y + C_{2} z & = & D_{2} \\ C_{3} z & = & D_{3} \end{array}$

It is very helpful if the values of A₁, B₂, C₃areequal to 1

What are row operations?

Row operations are used to make the linear equations simpler to solve

They do not affect the solution

You can switch any two rows

$[\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | \begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$ can be written as $[\begin{matrix} a_{3} & b_{3} & c_{3} \\ a_{2} & b_{2} & c_{2} \\ a_{1} & b_{1} & c_{1} \end{matrix} | \begin{matrix} d_{3} \\ d_{2} \\ d_{1} \end{matrix}]$ using r₁↔ r₃

This is useful for getting zeros to the bottom
Or getting a one to the top

You can multiply any row by a (non-zero) constant

$[\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | \begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$ can be written as $[\begin{matrix} a_{1} & b_{1} & c_{1} \\ k a_{2} & k b_{2} & k c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | \begin{matrix} d_{1} \\ k d_{2} \\ d_{3} \end{matrix}]$ using k×r₂ → r₂

This is useful for getting a 1 as the first non-zero value in a row

You can add multiples of a row to another row

$[\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | \begin{matrix} d_{1} \\ d_{2} \\ d_{3} \end{matrix}]$ can be written as $[\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} + 5 a_{3} & b_{2} + 5 b_{3} & c_{2} + 5 c_{3} \\ a_{3} & b_{3} & c_{3} \end{matrix} | \begin{matrix} d_{1} \\ d_{2} + 5 d_{3} \\ d_{3} \end{matrix}]$ using r₂+ 5r₃ → r₂

This is useful for creating zeros under a 1

How can I row reduce a system of linear equations?

STEP 1: Get a 1 in the top left corner
- You can do this by dividing the row by the current value in its place
- If the current value is 0 or an awkward number then you can swap rows first
  - $[\begin{matrix} 1 & B_{1} & C_{1} \\ * & * & * \\ * & * & * \end{matrix} | \begin{matrix} D_{1} \\ * \\ * \end{matrix}]$
STEP 2: Get 0’s in the entries below the 1
- You can do this by adding/subtracting a multiple of the first row to each row
  - $[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & * & * \\ 0 & * & * \end{matrix} | \begin{matrix} D_{1} \\ * \\ * \end{matrix}]$
STEP 3: Ignore the first row and column as they are now complete
- Repeat STEPS 1 - 2 to the remaining section
  - Get a 1: $[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & 1 & C_{2} \\ 0 & * & * \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ * \end{matrix}]$
  - Then 0 underneath: $[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & 1 & C_{2} \\ 0 & 0 & * \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ * \end{matrix}]$
STEP 4: Get a 1 in the third row
- Using the same idea as STEP 1
  - $[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & 1 & C_{2} \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ D_{3} \end{matrix}]$

How do I solve a system of linear equations once it is in row reduced form?

Once you row reduced the equations you can then convert back to the variables

$[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & 1 & C_{2} \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ D_{3} \end{matrix}]$ corresponds to $\begin{array}{rcl} x + B_{1} y + C_{1} z & = & D_{1} \\ y + C_{2} z & = & D_{2} \\ z & = & D_{3} \end{array}$

Solve the equations starting at the bottom

You have the value for z from the third equation
Substitute z into the second equation and solve for y
Substitute z and y into the first each and solve for x

Exam Tip

To reduce the number of operations you do whilst solving a system of operations, you can do a couple of things:
- You can set up your original matrix with the equations in any order, so if one of the equations already has a 1 in the top left corner, put that one first
- You do not need to make every equation so that it only has a single variable with a value of 1, you just need to do that for 1 of the equations and use that result to work out the others

Worked Example

Solve the following system of linear equations using algebra.

$\begin{array}{rcl} 2 x - 3 y + 4 z & = & 14 \\ x + 2 y - 2 z & = & - 2 \\ 3 x - y - 2 z & = & 10 \end{array}$

1-10-2-ib-aa-hl-row-reduction-we-solution

Number of Solutions to a System

How many solutions can a system of linear equations have?

There could be

1 unique solution
No solutions
An infinite number of solutions

You can determine the case by looking at the row reduced form

How do I know if the system of linear equations has no solutions?

Systems with no solutions are called inconsistent
When trying to solve the system after using the row reduction method you will end up with a mathematical statement which is never true:

Such as: 0 = 1

The row reduced system will contain:

At least one row where the entries to the left of the line are zero and the entry on the right of the line is non-zero

Such a row is called inconsistent

For example:

Row 2 is inconsistent $[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ D_{3} \end{matrix}]$ if D₂is non-zero

How do I know if the system of linear equations has infinite solutions?

Systems with at least one solution are called consistent
- The solution could be unique or there could be an infinite number of solutions
When trying to solve the system after using the row reduction method you will end up with a mathematical statement which is always true
- Such as: 0 = 0
The row reduced system will contain:
- At least one row where all the entries are zero
- No inconsistent rows
- For example:
  - $[\begin{matrix} 1 & B_{1} & C_{1} \\ 0 & 1 & C_{2} \\ 0 & 0 & 0 \end{matrix} | \begin{matrix} D_{1} \\ D_{2} \\ 0 \end{matrix}]$

How do I find the general solution of a dependent system?

A dependent system of linear equations is one where there are infinite number of solutions
The general solution will depend on one or two parameters
In the case where two rows are zero

Let the variables corresponding to the zero rows be equal to the parameters λ & μ

For example: If the first and second rows are zero rows then let x = λ & y = μ

Find the third variable in terms of the two parameters using the equation from the third row

For example: z = 4λ – 5μ + 6

In the case where only one row is zero

Let the variable corresponding to the zero row be equal to the parameter λ

For example: If the first row is a zero row then let x = λ

Find the remaining two variables in terms of the parameter using the equations formed by the other two rows

For example: y = 3λ – 5 & z = 7 – 2λ

Exam Tip

Common questions that pop up in an IB exam include questions with equations of lines
Being able to recognise whether there are no solutions, 1 solution or infinite solutions is really useful for identifying if lines are coincident, skew or intersect!

Worked Example

$\begin{array}{rcl} x + 2 y - z & = & 3 \\ 3 x + 7 y + z & = & 4 \\ x - 9 z & = & k \end{array}$

Given that the system of linear equations has an infinite number of equations, find the value of

k

1-10-2-ib-aa-hl-general-solution-a-we-solution

Find a general solution to the system.

1-10-2-ib-aa-hl-general-solution-b-we-solution

DP IB Maths: AA HL

Revision Notes

1.10.2 Algebraic Solutions

Row Reduction

How can I write a system of linear equations?

What is a row reduced system of linear equations?

What are row operations?

How can I row reduce a system of linear equations?

How do I solve a system of linear equations once it is in row reduced form?

Exam Tip

Worked Example

Number of Solutions to a System

How many solutions can a system of linear equations have?

How do I know if the system of linear equations has no solutions?

How do I know if the system of linear equations has infinite solutions?

How do I find the general solution of a dependent system?

Exam Tip

Worked Example

Author: Daniel

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