Solving three equations in three unknowns

2 Solving three equations in three unknowns

Cramer’s rule can be extended to larger systems of simultaneous equations but the calculational effort increases rapidly as the size of the system increases. We quote Cramer’s rule for a system of three equations.

Key Point 2

Cramer’s Rule for Three Equations The unique solution to the system of equations:

\begin{array}{rcl} a_{11} x_{1} + a_{12} x_{2} + a_{13} x_{3} & = & b_{1} \\ a_{21} x_{1} + a_{22} x_{2} + a_{23} x_{3} & = & b_{2} \\ a_{31} x_{1} + a_{32} x_{2} + a_{33} x_{3} & = & b_{3} \end{array}

x_{1} = \frac{Δ_{x_{1}}}{Δ}, x_{2} = \frac{Δ_{x_{2}}}{Δ}, x_{3} = \frac{Δ_{x_{3}}}{Δ}

in which

Δ = |\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}|

and

Δ_{x_{1}} = |\begin{matrix} b_{1} & a_{12} & a_{13} \\ b_{2} & a_{22} & a_{23} \\ b_{3} & a_{32} & a_{33} \end{matrix}| Δ_{x_{2}} = |\begin{matrix} a_{11} & b_{1} & a_{13} \\ a_{21} & b_{2} & a_{23} \\ a_{31} & b_{3} & a_{33} \end{matrix}| Δ_{x_{3}} = |\begin{matrix} a_{11} & a_{12} & b_{1} \\ a_{21} & a_{22} & b_{2} \\ a_{31} & a_{32} & b_{3} \end{matrix}|

If $Δ = 0$ this method of solution cannot be used.

Notice that the structure of the fractions is similar to that for the two-equation case. For example, the determinant forming the numerator of $x_{1}$ is obtained from the determinant of coefficients, $Δ$ , by replacing the first column by the right-hand sides of the equations.

Notice too the increase in calculation: in the two-equation case we had to evaluate three $2 \times 2$ determinants, whereas in the three-equation case we have to evaluate four $3 \times 3$ determinants. Hence Cramer’s rule is not really practicable for larger systems.

Task!

Use Cramer’s rule to solve the system

\begin{array}{rcl} x_{1} - 2 x_{2} + x_{3} & = & 3 \\ 2 x_{1} + x_{2} - x_{3} & = & 5 \\ 3 x_{1} - x_{2} + 2 x_{3} & = & 12 . \end{array}

First check that $Δ \neq 0$ :

$Δ = |\begin{matrix} 1 & - 2 & 1 \\ 2 & 1 & - 1 \\ 3 & - 1 & 2 \end{matrix}| .$

Expanding along the top row,

\begin{array}{rcl} Δ & = & 1 \times |\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix}| - (- 2) \times |\begin{matrix} 2 & - 1 \\ 3 & 2 \end{matrix}| + 1 \times |\begin{matrix} 2 & 1 \\ 3 & - 1 \end{matrix}| \\ = & 1 \times (2 - 1) + 2 \times (4 + 3) + 1 \times (- 2 - 3) \\ = & 1 + 14 - 5 = 10 \end{array}

Now find the value of $x_{1}$ . First write down the expression for $x_{1}$ in terms of determinants:

$x_{1} = |\begin{matrix} 3 & - 2 & 1 \\ 5 & 1 & - 1 \\ 12 & - 1 & 2 \end{matrix}| \div Δ$

Now calculate $x_{1}$ explicitly:

The numerator is found by expanding along the top row to be

\begin{array}{rcl} 3 & \times & |\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix}| - (- 2) \times |\begin{matrix} 5 & - 1 \\ 12 & 2 \end{matrix}| + 1 \times |\begin{matrix} 5 & 1 \\ 12 & - 1 \end{matrix}| \\ = & 3 \times 1 + 2 \times 22 + 1 \times (- 17) \\ = & 30 \end{array}

Hence $x_{1} = \frac{1}{10} \times 30 = 3$

In a similar way find the values of $x_{2}$ and $x_{3}$ :

\begin{array}{rcl} x_{2} & = & \frac{1}{10} |\begin{matrix} 1 & 3 & 1 \\ 2 & 5 & - 1 \\ 3 & 12 & 2 \end{matrix}| \\ = & \frac{1}{10} \{1 \times |\begin{matrix} 5 & - 1 \\ 12 & 2 \end{matrix}| - 3 \times |\begin{matrix} 2 & - 1 \\ 3 & 2 \end{matrix}| + 1 \times |\begin{matrix} 2 & 5 \\ 3 & 12 \end{matrix}|\} \\ = & \frac{1}{10} {22 - 3 \times 7 + 9} = 1 \\ x_{3} & = & \frac{1}{10} \{1 \times |\begin{matrix} 1 & 5 \\ - 1 & 12 \end{matrix}| - (- 2) \times |\begin{matrix} 2 & 5 \\ 3 & 12 \end{matrix}| + 3 \times |\begin{matrix} 2 & 1 \\ 3 & - 1 \end{matrix}|\} \\ = & \frac{1}{10} \{17 + 2 \times 9 + 3 \times (- 5)\} = 2 \end{array}

2 Solving three equations in three unknowns

Key Point 2

Task!

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