Solving system of simultaneous equations in 2 unknowns

Matrix solution of system of linear equations

There are two main methods of solving systems of linear equations using matrices: the inverse matrix method and Cramer's rule.

Solving system of linear equations by the inverse matrix method

A system of two equations with two unknowns

Consider the system of linear equations:

$$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$

In matrix form, this becomes:

$$\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$

Let $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$. The system can be written as:

$$A \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$

Pre-multiplying by $A^{-1}$:

$$A^{-1}A \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$

Since $A^{-1}A = I$ (the identity matrix):

$$\begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$

Example 10.17: A woman owns 21 domestic animals (goats and chickens) with a total of 76 legs. How many of each animal does she own?

Solution:

Let $x$ be the number of goats and $y$ be the number of chickens. Then:

$$\begin{cases} x + y = 21 \\ 4x + 2y = 76 \end{cases}$$

In matrix form:

$$\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 21 \\ 76 \end{bmatrix}$$

Let $A = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$. $\det(A) = (1 \cdot 2) - (1 \cdot 4) = -2$.

$A^{-1} = \frac{1}{-2} \begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix}$

$$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} 21 \\ 76 \end{bmatrix} = \begin{bmatrix} (-1)(21) + (\frac{1}{2})(76) \\ (2)(21) + (-\frac{1}{2})(76) \end{bmatrix} = \begin{bmatrix} 17 \\ 4 \end{bmatrix}$$

Therefore, $x = 17$ (goats) and $y = 4$ (chickens).

A system of three equations with three unknowns

Consider the system:

$$\begin{cases} a_{11}x + b_{11}y + c_{11}z = d_1 \\ a_{22}x + b_{22}y + c_{22}z = d_2 \\ a_{33}x + b_{33}y + c_{33}z = d_3 \end{cases}$$

In matrix form:

$$\begin{bmatrix} a_{11} & b_{11} & c_{11} \\ a_{22} & b_{22} & c_{22} \\ a_{33} & b_{33} & c_{33} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$$

Let $A$ be the coefficient matrix. Then:

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = A^{-1} \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}$$

The process involves finding the inverse of the 3x3 matrix $A$ (which involves finding minors, cofactors, the adjoint, and the determinant), and then multiplying it by the constant matrix.

Solving a system of linear equations using Cramer's rule

A system of two equations and two unknowns

Consider the system:

$$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$$

Cramer's rule states:

$$x = \frac{\begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} = \frac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1}, \quad y = \frac{\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}} = \frac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1}$$

Example 10.20: Solve:

$$\begin{cases} 2x + 3y = 9 \\ 4x - 2y = 10 \end{cases}$$

Solution:

$$x = \frac{\begin{vmatrix} 9 & 3 \\ 10 & -2 \end{vmatrix}}{\begin{vmatrix} 2 & 3 \\ 4 & -2 \end{vmatrix}} = \frac{-18 - 30}{-4 - 12} = \frac{-48}{-16} = 3$$ $$y = \frac{\begin{vmatrix} 2 & 9 \\ 4 & 10 \end{vmatrix}}{\begin{vmatrix} 2 & 3 \\ 4 & -2 \end{vmatrix}} = \frac{20 - 36}{-16} = \frac{-16}{-16} = 1$$

A system of three equations with three unknowns

For the system (10.17), Cramer's rule is:

$$x = \frac{|A_1|}{|A|}, \quad y = \frac{|A_2|}{|A|}, \quad z = \frac{|A_3|}{|A|}$$

Where $|A|$ is the determinant of the coefficient matrix, and $|A_1|$, $|A_2|$, and $|A_3|$ are the determinants of matrices formed by replacing the first, second, and third columns of $A$, respectively, with the constants $d_1$, $d_2$, and $d_3$.

Example 10.21: Solve the following system of equations using Cramer's rule:

$$\begin{cases} x + 2y + 3z = 7 \\ 2x + y - z = 7 \\ 3x + y + 2z = 0 \end{cases}$$

Solution:

In matrix form:

$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & -1 \\ 3 & 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 7 \\ 7 \\ 0 \end{bmatrix}$$

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & -1 \\ 3 & 1 & 2 \end{bmatrix}$. Calculate the determinant of $A$:

$|A| = 1\begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} - 2\begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} + 3\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 1(2+1) - 2(4+3) + 3(2-3) = 3 - 14 - 3 = -14$

Now, create the matrices $A_1$, $A_2$, and $A_3$ by replacing the columns of $A$ with the constant vector:

$$A_1 = \begin{bmatrix} 7 & 2 & 3 \\ 7 & 1 & -1 \\ 0 & 1 & 2 \end{bmatrix}, \quad A_2 = \begin{bmatrix} 1 & 7 & 3 \\ 2 & 7 & -1 \\ 3 & 0 & 2 \end{bmatrix}, \quad A_3 = \begin{bmatrix} 1 & 2 & 7 \\ 2 & 1 & 7 \\ 3 & 1 & 0 \end{bmatrix}$$

Calculate their determinants:

$|A_1| = 7\begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} - 2\begin{vmatrix} 7 & -1 \\ 0 & 2 \end{vmatrix} + 3\begin{vmatrix} 7 & 1 \\ 0 & 1 \end{vmatrix} = 7(3) - 2(14) + 3(7) = 21 - 28 + 21 = 14$

$|A_2| = 1\begin{vmatrix} 7 & -1 \\ 0 & 2 \end{vmatrix} - 7\begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} + 3\begin{vmatrix} 2 & 7 \\ 3 & 0 \end{vmatrix} = 1(14) - 7(7) + 3(-21) = 14 - 49 - 63 = -98$

$|A_3| = 1\begin{vmatrix} 1 & 7 \\ 1 & 0 \end{vmatrix} - 2\begin{vmatrix} 2 & 7 \\ 3 & 0 \end{vmatrix} + 7\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 1(-7) - 2(-21) + 7(-1) = -7 + 42 - 7 = 28$

Now, apply Cramer's rule:

$$x = \frac{|A_1|}{|A|} = \frac{14}{-14} = -1$$ $$y = \frac{|A_2|}{|A|} = \frac{-98}{-14} = 7$$ $$z = \frac{|A_3|}{|A|} = \frac{28}{-14} = -2$$

Therefore, the solution is $x = -1$, $y = 7$, and $z = -2$.