Crammer’s Rule (Solution Of Simultaneous Equations Using Determinant) - SS3 Mathematics Lesson Note

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TWO EQUATIONS IN TWO UNKNOWNS

Given two equations,

\[a_{11}x + a_{12}y = c_{1}\]

\(a_{12}x + a_{22}y = c_{2}\), where \(a_{11},\ \ a_{12},\ \ a_{21},\ \ a_{22},\ \ c_{1}\ \)and\(\ c_{2}\) are constants.

To solve for \(x\) and \(y\) using the determinant method, we put the coefficients in determinant form as follows:

Find the determinant \(\mathrm{\Delta} = \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right|\)

Then \(\mathrm{\Delta}x = \left| \begin{matrix} c_{1} & a_{12} \\ c_{2} & a_{22} \\ \end{matrix} \right|\) is obtained by replacing the column of \(x\):\(\begin{pmatrix} a_{11} \\ a_{12} \\ \end{pmatrix}\) by \(\begin{pmatrix} c_{1} \\ c_{2} \\ \end{pmatrix}\)

Then \(\mathrm{\Delta}y = \left| \begin{matrix} a_{11} & c_{1} \\ a_{21} & c_{2} \\ \end{matrix} \right|\) is obtained by replacing the column of \(y\): \(\begin{pmatrix} a_{12} \\ a_{22} \\ \end{pmatrix}\) by \(\begin{pmatrix} c_{1} \\ c_{2} \\ \end{pmatrix}\)

\[\mathrm{\Delta} = \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} \right| = a_{11}a_{22} - a_{12}a_{21}\]

\[\mathrm{\Delta}x = \left| \begin{matrix} c_{1} & a_{12} \\ c_{2} & a_{22} \\ \end{matrix} \right| = c_{1}a_{22} - a_{12}c_{2}\]

\[\mathrm{\Delta}y = \left| \begin{matrix} a_{11} & c_{1} \\ a_{21} & c_{2} \\ \end{matrix} \right| = a_{11}c_{2} - c_{1}a_{21}\]

\[x = \frac{\mathrm{\Delta}x}{\mathrm{\Delta}} = \frac{c_{1}a_{22} - a_{12}c_{2}}{a_{11}a_{22} - a_{12}a_{21}}\]

\[y = \frac{\mathrm{\Delta}y}{\mathrm{\Delta}} = \frac{a_{11}c_{2} - c_{1}a_{21}}{a_{11}a_{22} - a_{12}a_{21}}\]

Example 6 Use the determinant method to solve the simultaneous equations: \(x + 5y = - 3\) and \(2x - y = 5\)

Solution

\[\mathrm{\Delta} = \left| \begin{matrix} 1 & 5 \\ 2 & - 1 \\ \end{matrix} \right| = 1( - 1) - 5(2) = - 1 - 10 = - 11\]

\[\mathrm{\Delta}x = \left| \begin{matrix} - 3 & 5 \\ 5 & - 1 \\ \end{matrix} \right| = ( - 3)( - 1) - 5(5) = 3 - 25 = - 22\]

\[\mathrm{\Delta}y = \left| \begin{matrix} 1 & - 3 \\ 2 & 5 \\ \end{matrix} \right| = 1(5) - ( - 3)(2) = 5 + 6 = 11\]

\[x = \frac{\mathrm{\Delta}x}{\mathrm{\Delta}} = \frac{- 22}{- 11} = 2\]

\[y = \frac{\mathrm{\Delta}y}{\mathrm{\Delta}} = \frac{11}{- 11} = - 1\]

Thus, \(x = 2\) and \(y = - 1\)