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# IB Mathematics AA HL Flashcards- Systems of Liner Equations

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[h] IB Mathematics AA HL Flashcards- Systems of Liner Equations

[q] Systems of Liner Equations

[a]

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)

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SUBSTITUTION → $$y = 2x + 10$$

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

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

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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:

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

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

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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):

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

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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 → $$\frac{-1}{5}$$:

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

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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 → $$\frac{1}{2}$$:

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

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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:

$x + y + 2z = 1 x + z = 2 y + z = 0$

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

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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

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## IB Mathematics AA HL Flashcards- Systems of Liner Equations

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