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Exercises in solving simultaneous equations with 2 variables.

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Simultaneous Equations

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Solve the following simultaneous equations.

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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.

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

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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 $y$.

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For the first question:

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

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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.

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

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

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Rearranging for $A^{-1}C = b$:

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

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

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$\\simplify{{maA[0][0]}}x + \\simplify{{maA[0][1]}}y = \\simplify{{a1}}$

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

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$x =$ [[0]]

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$y =$ [[1]]

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$\\simplify{{maB[0][0]}}x + \\simplify{{maB[0][1]}}y = \\simplify{{b1}}$

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$\\simplify{{maB[1][0]}}x +\\simplify{{maB[1][1]}}y = \\simplify{{b2}}$

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$x =$ [[0]]

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$y =$ [[1]]

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