Simultaneous Equations

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Name	Status	Author	Last Modified
Simultaneous Equations	Ready to use	Mark Patterson	20/09/2020 19:01
Ugur's copy of Simultaneous Equations	Ready to use	Ugur Efem	12/01/2022 11:43

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Question statement

Solve the following simultaneous equations.

Give any introductory information the student needs.

			Name	Type	Generated Value

ax	integer	1
ay	integer	2
maA	matrix	matrix([2,-7],[7,2])
a1	number	-12
a2	number	11
bx	integer	4
by	integer	4
maB	matrix	matrix([6,6],[-4,8])
b1	number	48
b2	number	16
maAA	matrix	matrix([2,7],[-7,2])
detA	number	53

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xxxxxxxxxx
 
random(1..9)*random(1,1,1,-1)

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Description

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Generated value: `integer`

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Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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$\simplify{{maA[0][0]}}x + \simplify{{maA[0][1]}}y = \simplify{{a1}}$

$\simplify{{maA[1][0]}}x +\simplify{{maA[1][1]}}y = \simplify{{a2}}$

$x =$ Gap 0.

$y =$ Gap 1.

Advice

To solve simultaneous equations, first rewrite the equations in matrix form of $Ab = C$ , where the matrix $A$ will contain the coefficients of the equation, $b$ will contain the unknown values and $C$ will contain the constants.

We rearrange $Ab = C$ to get $A^{-1}C = b$ .

So, we need to find the inverse of $A$ and multiple it with $C$ . This will give us $b$ , i.e. the values for $x$ and $y$ .

For the first question:

$A = \simplify{{maA}}$ $b = \begin{pmatrix}x\\y\end{pmatrix}$ $C = \begin{pmatrix}\var{a1}\\ \var{a2}\end{pmatrix}$

To find the inverse of $A$ , switch the elements on the leading diagonal, change the signs on the non-leading diagonal, and divide all elements by the determinant.

The determinant of $A$ is $|A| = (\var{maA[0][0]}\times\var{maA[1][1]}) - (\var{maA[0][1]}\times\var{maA[1][0]}) = \var{detA}$

The inverse of $A$ is $A^{-1} = \dfrac{1}{\var{detA}}\times\begin{pmatrix} \var{maAA[0][0]} & \var{maAA[0][1]} \\ \var{maAA[1][0]} & \var{maAA[1][1]} \end{pmatrix} = \begin{pmatrix} \frac{\var{maAA[0][0]}}{\var{detA}} & \frac{\var{maAA[0][1]}}{\var{detA}} \\ \frac{\var{maAA[1][0]}}{\var{detA}} & \frac{\var{maAA[1][1]}}{\var{detA}} \end{pmatrix}$

Rearranging for $A^{-1}C = b$ :

$A^{-1}\times C = \begin{pmatrix} \frac{\var{maAA[0][0]}}{\var{detA}} & \frac{\var{maAA[0][1]}}{\var{detA}} \\ \frac{\var{maAA[1][0]}}{\var{detA}} & \frac{\var{maAA[1][1]}}{\var{detA}} \end{pmatrix} \times\begin{pmatrix}\var{a1}\\ \var{a2}\end{pmatrix} = \begin{pmatrix} \frac{\var{maAA[0][0]}}{\var{detA}}\times\var{a1} + \frac{\var{maAA[0][1]}}{\var{detA}}\times\var{a2} \\ \frac{\var{maAA[1][0]}}{\var{detA}}\times\var{a1} + \frac{\var{maAA[1][1]}}{\var{detA}}\times\var{a2} \end{pmatrix} = \begin{pmatrix}\var{ax}\\ \var{ay}\end{pmatrix}$

Therefore $x = \var{ax}$ and $y = \var{ay}$ .

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Simultaneous Equations

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Mark Patterson 4 years, 7 months ago

Mark Patterson 4 years, 7 months ago

Mark Patterson 4 years, 7 months ago