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Semi-worked example of solving simultaneous equations using matrices. Equation values are randomly generated. The student is walked through the steps needed to solve the equations.

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Solving Simultaneous Equations using Matrices

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Simultaneous equations consist of two equations with two unknown values and both equations must be solved at the same time. This can be done using matrices.

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Consider the following set of simultaneous equations:

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\\begin{align}
\\simplify{{mA[0][0]}x + {mA[0][1]}y} &= \\var{mC[0][0]} \\\\
\\simplify{{mA[1][0]}x + {mA[1][1]}y} &= \\var{mC[1][0]}
\\end{align}

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We can solve these equations using the following steps:

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(Input all numbers as fractions or integers and not as decimals.)

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value for x

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value for y

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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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$\\mathbf{A} = $ [[0]]    $\\mathbf{b} = $ \n
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\n\n\n\n\n\n\n\n\n\n
[[1]]
[[2]]
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\n       $\\mathbf{C} = $ [[3]]
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The coefficients are the values $x$ and $y$ are multiplied by. Take the number of $x$s in the first equation and place it in the first element of matrix $A$, $a_{11}$. Take the number of $y$s in the first equation and place it in the second element of matrix $A$, $a_{12}$ (first row, second column).

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Repeat for the second equation, but place coefficients into the second row elements of matrix $A$.

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The unknown values $x$ and $y$ go into matrix $b$, $x$ into $b_{11}$ and $y$ into $b_{21}$. We don't know these values yet, so we use $x$ and $y$ for now.

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The constants are the answers to the equations (the number to the right of the $=$ sign). The answer to the first equation goes into the first element of matrix $C$, $c_{11}$. The answer to the second equation goes into the second element of matrix $C$, $c_{21}$.

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Second, we find the inverse of the coefficient matrix $A$

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Switch the elements on the leading/main diagonal.

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Change the sign of the elelments on the secondary diagonal.

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i.e. $\\begin{pmatrix} d & -b\\\\ -c & a \\end{pmatrix}$

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Calculate the determinant of $A$

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i.e. $ad-bc$

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Divide every element in $A'$ by the determinant. (Enter answer as fractions.)

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Finally, find $A^{-1}C$. i.e multiply the inverse of $A$, $A^{-1}$, by $C$

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We have matrix $A^{-1} = \\begin{pmatrix} a_{11} & a_{12}\\\\ a_{21} & a_{22} \\end{pmatrix}$ multiplied by matrix $C = \\begin{pmatrix} c_{11}\\\\ c_{21} \\end{pmatrix}$.

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We know we can multiply these as the number of columns in $A^{-1}$, 2, is equal to the number of rows in $C$, 2.

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[2 x 2] x [2 x 1]

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We can also determine the order of the answer matrix:-

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To work out the values for the answer matrix we multiply:-

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$a'c_{11} = (a'_{11}\\times c_{11}) + (a'_{12}\\times c_{21}) =$ ?

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$a'c_{21} = (a'_{21}\\times c_{11}) + (a'_{22}\\times c_{21}) =$ ?

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Therefore $A^{-1}C = \\begin{pmatrix} a'c_{11}\\\\ a'c_{21} \\end{pmatrix}$ is?

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We started with the matrices $Ab=C$, and we have rearranged this formula to $A^{-1}C=b$ in order to find $b$. Therefore, $x$ and $y$ are?

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$x = $ [[0]]

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$y = $ [[1]]

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