By the end of this chapter, you will be able to:
Mastering these skills will help you solve practical problems and make smart decisions in the workplace!
In procurement management, handling data in matrix form is common when analyzing supplier performance, cost comparisons, or bid evaluations. A 2×2 matrix is a simple yet powerful tool to organize and solve such data efficiently. Solving a 2×2 matrix involves finding unknown variables from matrix equations, which is essential for decision-making in procurement contracts and negotiations.
A 2×2 matrix equation typically takes the form \( AX = B \), where \( A \) and \( B \) are known matrices and \( X \) is the unknown matrix to be solved. The matrix \( A \) is of order 2×2, meaning it has 2 rows and 2 columns, similarly for \( B \) and \( X \).
The general form is:
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} e \\ f \end{bmatrix} $$
where \(a, b, c, d, e, f\) are known numbers, and \(x, y\) are unknowns to be solved.
Example 1:
Find \(x\) and \(y\) if
$$ \begin{bmatrix} 3 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 18 \\ 7 \end{bmatrix} $$
Given: \(a=3, b=4, c=2, d=1, e=18, f=7\)
Step 1: Write the matrix equations:
$$ 3x + 4y = 18 $$
$$ 2x + y = 7 $$
Step 2: Solve one equation for one variable, e.g., from the second:
$$ y = 7 - 2x $$
Step 3: Substitute into the first:
$$ 3x + 4(7 - 2x) = 18 $$
$$ 3x + 28 - 8x = 18 $$
$$ -5x + 28 = 18 $$
$$ -5x = 18 - 28 = -10 $$
$$ x = \frac{-10}{-5} = 2 $$
Step 4: Substitute back to find \(y\):
$$ y = 7 - 2(2) = 7 - 4 = 3 $$
Answer: \(x = 2, y = 3\)
Example 2:
Solve for \(x\) and \(y\):
$$ \begin{bmatrix} 5 & -2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 16 \\ 13 \end{bmatrix} $$
Given: \(a=5, b=-2, c=1, d=3, e=16, f=13\)
Step 1: Write equations:
$$ 5x - 2y = 16 $$
$$ x + 3y = 13 $$
Step 2: Solve second for \(x\):
$$ x = 13 - 3y $$
Step 3: Substitute into first:
$$ 5(13 - 3y) - 2y = 16 $$
$$ 65 - 15y - 2y = 16 $$
$$ 65 - 17y = 16 $$
$$ -17y = 16 - 65 = -49 $$
$$ y = \frac{-49}{-17} = \frac{49}{17} \approx 2.88 $$
Step 4: Find \(x\):
$$ x = 13 - 3(2.88) = 13 - 8.64 = 4.36 $$
Answer: \(x \approx 4.36, y \approx 2.88\)
Example 3:
Find \(x\) and \(y\):
$$ \begin{bmatrix} 7 & 5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 41 \\ -1 \end{bmatrix} $$
Given: \(a=7, b=5, c=-3, d=2, e=41, f=-1\)
Step 1: Equations:
$$ 7x + 5y = 41 $$
$$ -3x + 2y = -1 $$
Step 2: Solve second for \(y\):
$$ 2y = -1 + 3x $$
$$ y = \frac{-1 + 3x}{2} $$
Step 3: Substitute into first:
$$ 7x + 5 \times \frac{-1 + 3x}{2} = 41 $$
$$ 7x + \frac{5(-1 + 3x)}{2} = 41 $$
Multiply both sides by 2 to clear denominator:
$$ 2 \times 7x + 5(-1 + 3x) = 82 $$
$$ 14x + 5(-1) + 15x = 82 $$
$$ 14x - 5 + 15x = 82 $$
$$ 29x - 5 = 82 $$
$$ 29x = 87 $$
$$ x = \frac{87}{29} = 3 $$
Step 4: Find \(y\):
$$ y = \frac{-1 + 3(3)}{2} = \frac{-1 + 9}{2} = \frac{8}{2} = 4 $$
Answer: \(x = 3, y = 4\)
The inverse matrix method is a faster approach to solve \( AX = B \) when matrix \(A\) is invertible. The solution is:
$$ X = A^{-1} B $$
where \(A^{-1}\) is the inverse of matrix \(A\). For a 2×2 matrix
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
the inverse is
$$ A^{-1} = \frac{1}{ad, bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$
provided \(ad, bc eq 0\).
Example 1:
Solve for \(X\) where
$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 11 \end{bmatrix} $$
Given: \(a=1, b=2, c=3, d=4\)
Step 1: Calculate determinant:
$$ ad, bc = (1)(4) - (2)(3) = 4 - 6 = -2 $$
Step 2: Find inverse:
$$ A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} $$
Step 3: Multiply \(A^{-1}\) by \(B\):
$$ X = A^{-1} B = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \begin{bmatrix} 5 \\ 11 \end{bmatrix} $$
Calculate first element:
$$ (-2)(5) + (1)(11) = -10 + 11 = 1 $$
Calculate second element:
$$ (1.5)(5) + (-0.5)(11) = 7.5 - 5.5 = 2 $$
Answer:
$$ X = \begin{bmatrix} 1 \\ 2 \end{bmatrix} $$
Example 2:
Solve
$$ A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}, \quad B = \begin{bmatrix} 30 \\ 18 \end{bmatrix} $$
Step 1: Determinant:
$$ ad, bc = (4)(6) - (7)(2) = 24 - 14 = 10 $$
Step 2: Inverse:
$$ A^{-1} = \frac{1}{10} \begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix} $$
Step 3: Multiply by \(B\):
$$ X = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix} \begin{bmatrix} 30 \\ 18 \end{bmatrix} $$
Calculate first element:
$$ (0.6)(30) + (-0.7)(18) = 18 - 12.6 = 5.4 $$
Calculate second element:
$$ (-0.2)(30) + (0.4)(18) = -6 + 7.2 = 1.2 $$
Answer:
$$ X = \begin{bmatrix} 5.4 \\ 1.2 \end{bmatrix} $$
Example 3:
Solve
$$ A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ 8 \end{bmatrix} $$
Step 1: Determinant:
$$ ad, bc = (2)(3) - (5)(1) = 6 - 5 = 1 $$
Step 2: Inverse:
$$ A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} $$
Step 3: Multiply by \(B\):
$$ X = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 20 \\ 8 \end{bmatrix} $$
Calculate first element:
$$ (3)(20) + (-5)(8) = 60 - 40 = 20 $$
Calculate second element:
$$ (-1)(20) + (2)(8) = -20 + 16 = -4 $$
Answer:
$$ X = \begin{bmatrix} 20 \\ -4 \end{bmatrix} $$
Matrix solutions help procurement professionals resolve systems such as price and quantity negotiations, supplier rating scores, and resource allocations. For example, if bids from two suppliers relate to two criteria, solving a matrix equation can determine optimal quantities or weighted scores.
Example 1:
A procurement officer has two suppliers with unit costs and delivery times represented as:
$$ \begin{bmatrix} 10 & 5 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 200 \\ 54 \end{bmatrix} $$
Where \(x\) and \(y\) represent quantities ordered from supplier 1 and 2 respectively. Solve for \(x\) and \(y\).
Step 1: Equations:
$$ 10x + 5y = 200 $$
$$ 2x + 3y = 54 $$
Step 2: Solve second for \(x\):
$$ 2x = 54 - 3y $$
$$ x = \frac{54 - 3y}{2} $$
Step 3: Substitute into first:
$$ 10 \times \frac{54 - 3y}{2} + 5y = 200 $$
Multiply both sides by 2:
$$ 10(54 - 3y) + 10y = 400 $$
$$ 540 - 30y + 10y = 400 $$
$$ 540 - 20y = 400 $$
$$ -20y = 400 - 540 = -140 $$
$$ y = \frac{-140}{-20} = 7 $$
Step 4: Find \(x\):
$$ x = \frac{54 - 3(7)}{2} = \frac{54 - 21}{2} = \frac{33}{2} = 16.5 $$
Answer:
$$ x = 16.5 \text{ units}, \quad y = 7 \text{ units} $$
Example 2:
A county government office evaluates two contractors on cost and time scores:
$$ \begin{bmatrix} 4 & 6 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 56 \\ 43 \end{bmatrix} $$
Find \(x\) and \(y\).
Step 1: Equations:
$$ 4x + 6y = 56 $$
$$ 3x + 5y = 43 $$
Step 2: Solve second for \(x\):
$$ 3x = 43 - 5y $$
$$ x = \frac{43 - 5y}{3} $$
Step 3: Substitute into first:
$$ 4 \times \frac{43 - 5y}{3} + 6y = 56 $$
Multiply both sides by 3:
$$ 4(43 - 5y) + 18y = 168 $$
$$ 172 - 20y + 18y = 168 $$
$$ 172 - 2y = 168 $$
$$ -2y = 168 - 172 = -4 $$
$$ y = 2 $$
Step 4: Find \(x\):
$$ x = \frac{43 - 5(2)}{3} = \frac{43 - 10}{3} = \frac{33}{3} = 11 $$
Answer:
$$ x = 11, \quad y = 2 $$
Example 3:
In a retail business, the demand and supply constraints are:
$$ \begin{bmatrix} 7 & 3 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 65 \\ 58 \end{bmatrix} $$
Find \(x\) and \(y\).
Step 1: Equations:
$$ 7x + 3y = 65 $$
$$ 4x + 5y = 58 $$
Step 2: Solve second for \(x\):
$$ 4x = 58 - 5y $$
$$ x = \frac{58 - 5y}{4} $$
Step 3: Substitute into first:
$$ 7 \times \frac{58 - 5y}{4} + 3y = 65 $$
Multiply both sides by 4:
$$ 7(58 - 5y) + 12y = 260 $$
$$ 406 - 35y + 12y = 260 $$
$$ 406 - 23y = 260 $$
$$ -23y = 260 - 406 = -146 $$
$$ y = \frac{-146}{-23} = \frac{146}{23} \approx 6.35 $$
Step 4: Find \(x\):
$$ x = \frac{58 - 5(6.35)}{4} = \frac{58 - 31.75}{4} = \frac{26.25}{4} = 6.56 $$
Answer:
$$ x \approx 6.56, \quad y \approx 6.35 $$
Cramer's Rule provides a direct method to solve 2×2 linear systems using determinants. For the system:
$$ \begin{cases} ax + by = e \\ cx + dy = f \end{cases} $$
the solution is:
$$ x = \frac{\det A_x}{\det A}, \quad y = \frac{\det A_y}{\det A} $$
where
$$ \det A = ad, bc $$
$$ \det A_x = \begin{vmatrix} e & b \\ f & d \end{vmatrix} = ed, bf $$
$$ \det A_y = \begin{vmatrix} a & e \\ c & f \end{vmatrix} = af, ce $$
Example 1:
Solve
$$ \begin{cases} 2x + 3y = 13 \\ 4x + y = 11 \end{cases} $$
Given: \(a=2, b=3, c=4, d=1, e=13, f=11\)
Step 1: Calculate determinants:
$$ \det A = (2)(1) - (3)(4) = 2 - 12 = -10 $$
$$ \det A_x = (13)(1) - (3)(11) = 13 - 33 = -20 $$
$$ \det A_y = (2)(11) - (4)(13) = 22 - 52 = -30 $$
Step 2: Calculate \(x\) and \(y\):
$$ x = \frac{-20}{-10} = 2 $$
$$ y = \frac{-30}{-10} = 3 $$
Answer: \(x=2, y=3\)
Example 2:
Solve
$$ \begin{cases} 5x - 2y = 16 \\ 3x + 4y = 25 \end{cases} $$
Given: \(a=5, b=-2, c=3, d=4, e=16, f=25\)
Step 1: Determinants:
$$ \det A = (5)(4) - (-2)(3) = 20 + 6 = 26 $$
$$ \det A_x = (16)(4) - (-2)(25) = 64 + 50 = 114 $$
$$ \det A_y = (5)(25) - (3)(16) = 125 - 48 = 77 $$
Step 2: Values:
$$ x = \frac{114}{26} \approx 4.38 $$
$$ y = \frac{77}{26} \approx 2.96 $$
Answer: \(x \approx 4.38, y \approx 2.96\)
Example 3:
Solve
$$ \begin{cases} 7x + 3y = 31 \\ 2x + 5y = 24 \end{cases} $$
Given: \(a=7, b=3, c=2, d=5, e=31, f=24\)
Step 1: Determinants:
$$ \det A = (7)(5) - (3)(2) = 35 - 6 = 29 $$
$$ \det A_x = (31)(5) - (3)(24) = 155 - 72 = 83 $$
$$ \det A_y = (7)(24) - (2)(31) = 168 - 62 = 106 $$
Step 2: Values:
$$ x = \frac{83}{29} \approx 2.86 $$
$$ y = \frac{106}{29} \approx 3.66 $$
Answer: \(x \approx 2.86, y \approx 3.66\)
\[ \begin{bmatrix} 6 & 2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 20 \\ 13 \end{bmatrix} \] (5 marks)
\[ \begin{bmatrix} 3 & 5 \\ 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 19 \\ 31 \end{bmatrix} \] (6 marks)
\[ \begin{cases} 8x + 4y = 40 \\ 3x + 7y = 43 \end{cases} \]
Find \(x\) and \(y\) using Cramer's Rule. (6 marks)
\[ \begin{bmatrix} 4 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 11 \\ 16 \end{bmatrix} \]
Solve for \(x\) and \(y\). (5 marks)
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter introduced the fundamental concept of matrices, emphasizing their structure and significance in various calculations. It covered essential matrix operations including addition, subtraction, division, and multiplication, detailing how these operations are performed and their mathematical rules. The chapter then focused on solving 2 by 2 matrix equations, providing methods to handle such systems effectively. Determinants of 2 by 2 matrices were explained as a critical value for matrix properties and solution existence. The concept of the inverse of a 2 by 2 matrix was explored, showing how to compute it and why it is important for solving linear equations. Finally, the chapter demonstrated practical applications of matrices in solving business operations, highlighting their usefulness in real-world problem solving and decision-making processes.
Given two matrices representing quantities of goods ordered from two suppliers for three products:
\( A = \begin{bmatrix} 10 & 15 & 20 \\ 5 & 10 & 15 \end{bmatrix} \),
\( B = \begin{bmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \end{bmatrix} \),
find the matrix \( C = A + B \) representing total quantities ordered. (2 marks)
A procurement officer recorded the unit prices of three items from two suppliers as:
\( P = \begin{bmatrix} 150 & 200 & 250 \\ 140 & 210 & 240 \end{bmatrix} \) Ksh,
and discounts offered as:
\( D = \begin{bmatrix} 10 & 15 & 20 \\ 5 & 10 & 15 \end{bmatrix} \) Ksh.
Calculate the matrix \( R = P - D \) representing the effective prices after discount. (2 marks)
A procurement officer at a county government has two matrices representing the quantities of two types of supplies requested from two suppliers:
\( A = \begin{bmatrix} 50 & 30 \\ 20 & 40 \end{bmatrix} \), \( B = \begin{bmatrix} 10 & 15 \\ 25 & 35 \end{bmatrix} \).
Calculate \( A + B \). (4 marks)
Given matrices \( C = \begin{bmatrix} 70 & 60 \\ 55 & 45 \end{bmatrix} \) and \( D = \begin{bmatrix} 20 & 15 \\ 10 & 5 \end{bmatrix} \), find \( C - D \). (4 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.