IB Mathematics AA HL Flashcards- Systems of Liner Equations

Whilst students have been solving systems of 2 equations for years, now we need to solve 3 equations with 3 variables.

[q]

TWO VARIABLES
A reminder of methods:

E.G. 1/4 Solve:
\[
\begin{aligned}
2x + 3y &= 6 \quad (1) \\
2x – y &= -10 \quad (2)
\end{aligned}
\]

[a]

ELIMINATION →
\[
0 – (2)(0x) + 4y = 16 \\
y = 4
\]
\[
\text{Plug into (2)} 2x + 3(4) = 6 \\
2x = -6 \\
x = -3
\]
Intersection: (-3, 4)

[q]

SUBSTITUTION → \( y = 2x + 10 \)

\[
\text{Plug } y \text{ into (1)}: 2x + 3(2x+10) = 6 \\
8x + 30 = 6 \\
8x = -24 \\
x = -3
\]

\[
\text{Plug into (2)}: 2(-3) + 3y = 6 \\
-6 + 3y = 6 \Rightarrow 3y = 12 \Rightarrow y = 4
\]

[a]

Intersection: (-3, 4)

GRAPHICAL →
(Graph of two lines showing an intersection at (-3, 4))

[q]
THREE VARIABLES
This method will essentially use a version of elimination. You can multiply any equation by a constant, switch equations, or add multiples of any equation to others. It will be displayed in regular equation form, or in ‘matrix’ form.

[a]

It should be clear to see that a solution has been reached when you get the matrix in the form:

\[
\left( \begin{array}{ccc|c}
1 & 0 & 0 & x \\
0 & 1 & 0 & y \\
0 & 0 & 1 & z \\
\end{array} \right)
\]

[q]

E.G. 2/4 Solve:

EQUATION
\[
\begin{aligned}
2x + y – z &= 2 \quad (1) \\
x + 3y + 2z &= 1 \quad (2) \\
2x + 4y + 6z &= 6 \quad (3)
\end{aligned}
\]

[a]

MATRIX →
\[
\left( \begin{array}{ccc|c}
2 & 1 & -1 & 2 \\
1 & 3 & 2 & 1 \\
2 & 4 & 6 & 6 \\
\end{array} \right)
\]

[q]
Switch (1) & (2):

\[
\left( \begin{array}{ccc|c}
1 & 3 & 2 & 1 \\
2 & 1 & -1 & 2 \\
2 & 4 & 6 & 6 \\
\end{array} \right)
\]

[a]
Switch row (1) & (2):

\[
\left( \begin{array}{ccc|c}
1 & 3 & 2 & 1 \\
1 & 1 & 1 & 1 \\
2 & 4 & 6 & 6 \\
\end{array} \right)
\]

[q]
Row (3) → ½:

\[
\left( \begin{array}{ccc|c}
1 & 3 & 2 & 1 \\
1 & 1 & 1 & 1 \\
1 & 2 & 3 & 3 \\
\end{array} \right)
\]

[a]
2 – 2x (3-1):

\[
\left( \begin{array}{ccc|c}
1 & 3 & 2 & 1 \\
0 & -5 & -1 & 2 \\
0 & -1 & 1 & 2 \\
\end{array} \right)
\]

[q]

ORDER OF PRIORITIES FOR MATRICES →
– A: Get a ‘1’ in the top-left corner
– B: Get \( 1 \ \ 0 \ \ 0 \) on the left

\[
\frac{2 + 3}{3}
\]
x + 3y + z = 1 \( \ \ \ \ \ \ \ \ \ \ (1) \)
y + z = 0 \( \ \ \ \ \ \ \ \ \ \ (2) \)
-y + z = 3 \( \ \ \ \ \ \ \ \ \ \ (3) \)

[a]

Row 2 → \(\frac{-1}{5}\):

\[
\left( \begin{array}{ccc|c}
1 & 3 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & -1 & 1 & 3 \\
\end{array} \right)
\]

[q]

C: Get a ‘1’ in the middle

\[
\left( \begin{array}{ccc|c}
1 & 3 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 2 & 2 \\
\end{array} \right)
\]

D: Get \( 0 \) in the middle column

[a]

Row 3 → \(\frac{1}{2}\):

\[
\left( \begin{array}{ccc|c}
1 & 3 & 1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
\end{array} \right)
\]

[q]

E: Get a ‘1’ in the bottom-right corner

\[
1 + 3(2) + 2 – 3
\]

\[
x = 2, \quad y = -1, \quad z = 1
\]

Our three equations/rows gives us the solution:
\(x = 2, y = -1, z = 1\).

OTHER CASES

There can also be no solutions, or infinite solutions.

[a]

E.G. 3/4 Solve:

\[
x + y + 2z = 1
x + z = 2
y + z = 0
\]

[q]

MATRIX →
\[
\left( \begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
1 & 0 & 1 & 2 \\
0 & 1 & 1 & 0 \\
\end{array} \right)
\]

[a]

\[
\left( \begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
0 & -1 & -1 & 1 \\
0 & 0 & 0 & 1 \\
\end{array} \right)
\]

→ If we stop to analyse the 3rd row, we see that we have \( 0x + 0y + 0z = 1 \), which isn’t possible.
The system is inconsistent → it has 0 solutions.

[q]

E.G. 4/4 Solve:

\[
2x + y – z = 4 \\
x + 3y + 7z = 7 \\
2x + 4y + 8z = 10
\]

[a]

MATRIX →
\[
\left( \begin{array}{ccc|c}
2 & 1 & -1 & 4 \\
1 & 3 & 7 & 7 \\
2 & 4 & 8 & 10 \\
\end{array} \right)
\]

[q]

Switch \( 0 \rightarrow 2 \):

\[
\left( \begin{array}{ccc|c}
1 & 1 & 1 & 7 \\
0 & -15 & -10 & -2 \\
0 & 3 & 4 & 6 \\
\end{array} \right)
\]

[a]

→ Here, we have all zeros on the third row. That just leaves us with:
\(x – z = 1\)
\(y + 3z = 2\), meaning for each value of \(z\), we get a new \(x\) & \(y\) value. If we set \(z = 0\), we get \( (1,2,0) \), if \(z = 2\),

[q]

we have \( (5,-4,2) \). In general, the solution is \( (1 + 2t, 2 – 3t, t) \) → an infinity of solutions.

[a]

NOTE: You can use the ‘rref’ function on your GDC to do the whole matrix process and find regular solutions to these problems.

[x] Exit text