For matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is:

$\det(A) = ad - bc$

The determinant tells us if a unique solution exists. If $\det(A) = 0$, the equations are either dependent (infinitely many solutions) or inconsistent (no solution).

If $A$ has a non-zero determinant, we can multiply both sides of $A\mathbf{x} = \mathbf{c}$ by $A^{-1}$:

$A^{-1}A\mathbf{x} = A^{-1}\mathbf{c}$

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

$\mathbf{x} = A^{-1}\mathbf{c}$

Finding the Inverse of a 2×2 Matrix

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $\det(A) \neq 0$:

$A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Worked Example

Solve using the matrix inversion method:

Solution:

Step 1: Write in matrix form: $A = \begin{pmatrix} 2 & 1 \\ 3 & -2 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 10 \\ 1 \end{pmatrix}$

Step 2: Find the determinant: $\det(A) = (2)(-2) - (1)(3) = -4 - 3 = -7$

Since $\det(A) \neq 0$, we can proceed.

Step 3: Find the inverse: $A^{-1} = \frac{1}{-7} \begin{pmatrix} -2 & -1 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} \frac{2}{7} & \frac{1}{7} \\ \frac{3}{7} & -\frac{2}{7} \end{pmatrix}$

Step 4: Multiply by the constant vector: $\mathbf{x} = A^{-1}\mathbf{c} = \begin{pmatrix} \frac{2}{7} & \frac{1}{7} \\ \frac{3}{7} & -\frac{2}{7} \end{pmatrix} \begin{pmatrix} 10 \\ 1 \end{pmatrix}$

$x = \frac{2}{7}(10) + \frac{1}{7}(1) = \frac{20}{7} + \frac{1}{7} = \frac{21}{7} = 3$

$y = \frac{3}{7}(10) - \frac{2}{7}(1) = \frac{30}{7} - \frac{2}{7} = \frac{28}{7} = 4$

Answer: $x = 3, \quad y = 4$

Cramer's rule uses determinants directly to find each variable.

Given:

Let:

• $D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1$ (determinant of coefficient matrix)
• $D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1b_2 - c_2b_1$ (replace first column with constants)
• $D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1c_2 - a_2c_1$ (replace second column with constants)

Then: $x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}$

Worked Example

Solve using Cramer's rule:

Solution:

Step 1: Find the main determinant $D$: $D = \begin{vmatrix} 6 & 1 \\ 3 & 1 \end{vmatrix} = (6)(1) - (3)(1) = 6 - 3 = 3$

Step 2: Find $D_x$ (replace first column with constants): $D_x = \begin{vmatrix} 15 & 1 \\ 9 & 1 \end{vmatrix} = (15)(1) - (9)(1) = 15 - 9 = 6$

Step 3: Find $D_y$ (replace second column with constants): $D_y = \begin{vmatrix} 6 & 15 \\ 3 & 9 \end{vmatrix} = (6)(9) - (3)(15) = 54 - 45 = 9$

Step 4: Compute each variable: $x = \frac{D_x}{D} = \frac{6}{3} = 2$ $y = \frac{D_y}{D} = \frac{9}{3} = 3$

Answer: $x = 2, \quad y = 3$

If $D = 0$:

• And $D_x = 0$ and $D_y = 0$: infinitely many solutions (dependent equations)
• And at least one of $D_x$ or $D_y$ is non-zero: no solution (inconsistent equations)

In such cases, matrix methods cannot give a unique solution.

In Tanzania, market vendors often need to calculate how many kilograms of different crops (like maize and beans) they can buy with a fixed amount of money when they know the total weight and the price per kilogram. A shopkeeper in Dar es Salaam might have two equations representing the total cost and total weight of two types of rice, and using matrix methods (matrix inversion or Cramer's rule), they can quickly determine the exact quantity of each type to purchase. This same method is used in engineering and economics to solve systems of linear equations that model supply and demand, mixing problems, and traffic flow analysis.