The oldest known simultaneous equations solved using something we would recognize as matrix methods are these (from the Chinese document The nine chapters on the mathematical arts circa 200 BCE ). The first problem of Chapter 8 is: 3 bundles of high-grade grain, 2 bundles of medium-grade grain and 1 bundle of low- grade grain give 39 units. 2 bundles of high-grade grain, 3 bundles of medium-grade grain, and I bundle of low-grade grain give 34 units. 1 bundle of high-grade grain, 2 bundles of medium-grade grain, and 3 bundles of low-grade grain yields 26 units. How many units does one bundle of each grain yield?

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To solve the given system of simultaneous equations, we'll determine how many units each bundle of high-grade grain (h), medium-grade grain (m), and low-grade grain (l) yields. Here's a step-by-step solution using matrix methods and algebraic manipulation.

## Problem Statement

We have the following system of equations based on the problem:

1. **Equation 1:**  
   $$ 3h + 2m + l = 39 $$

2. **Equation 2:**  
   $$ 2h + 3m + l = 34 $$

3. **Equation 3:**  
   $$ h + 2m + 3l = 26 $$

Our goal is to find the values of $$ h $$, $$ m $$, and $$ l $$.

## Step 1: Representing the System as a Matrix

First, let's represent the system in matrix form $$ A\mathbf{x} = \mathbf{b} $$, where:

$$
A = \begin{bmatrix}
3 & 2 & 1 \
2 & 3 & 1 \
1 & 2 & 3 \
\end{bmatrix}, \quad
\mathbf{x} = \begin{bmatrix}
h \
m \
l \
\end{bmatrix}, \quad
\mathbf{b} = \begin{bmatrix}
39 \
34 \
26 \
\end{bmatrix}
$$

## Step 2: Utilizing the Elimination Method

To solve the system, we'll use the elimination method to reduce the system to two equations with two variables.

### 2.1 Subtract Equation 2 from Equation 1:

$$
(3h + 2m + l) - (2h + 3m + l) = 39 - 34 \
3h - 2h + 2m - 3m + l - l = 5 \
h - m = 5 \quad \Rightarrow \quad h = m + 5 \quad \text{[Equation 4]}
$$

### 2.2 Substitute Equation 4 into Equation 1:

$$
3(m + 5) + 2m + l = 39 \
3m + 15 + 2m + l = 39 \
5m + l = 24 \quad \text{[Equation 5]}
$$

### 2.3 Substitute Equation 4 into Equation 2:

$$
2(m + 5) + 3m + l = 34 \
2m + 10 + 3m + l = 34 \
5m + l = 24 \quad \text{[Same as Equation 5]}
$$

### 2.4 Substitute Equation 4 into Equation 3:

$$
(m + 5) + 2m + 3l = 26 \
m + 5 + 2m + 3l = 26 \
3m + 3l = 21 \quad \Rightarrow \quad m + l = 7 \quad \text{[Equation 6]}
$$

Now, we have two equations with two variables:

1. **Equation 5:**  
   $$ 5m + l = 24 $$

2. **Equation 6:**  
   $$ m + l = 7 $$

## Step 3: Solving for $$ m $$ and $$ l $$

### 3.1 Subtract Equation 6 from Equation 5:

$$
(5m + l) - (m + l) = 24 - 7 \
4m = 17 \
m = \frac{17}{4} = 4.25
$$

### 3.2 Solve for $$ l $$ Using Equation 6:

$$
4.25 + l = 7 \
l = 7 - 4.25 \
l = 2.75
$$

### 3.3 Solve for $$ h $$ Using Equation 4:

$$
h = m + 5 = 4.25 + 5 = 9.25
$$

## Step 4: Conclusion

The values for each type of grain bundle are:

- **High-grade grain (h):** $$ 9.25 $$ units
- **Medium-grade grain (m):** $$ 4.25 $$ units
- **Low-grade grain (l):** $$ 2.75 $$ units

These values satisfy all three original equations, ensuring the solution is correct.

## Verification

Let's verify by plugging the values back into all three equations.

1. **Equation 1:**  
   $$ 3(9.25) + 2(4.25) + 2.75 = 27.75 + 8.5 + 2.75 = 39 $$ ✔️

2. **Equation 2:**  
   $$ 2(9.25) + 3(4.25) + 2.75 = 18.5 + 12.75 + 2.75 = 34 $$ ✔️

3. **Equation 3:**  
   $$ 9.25 + 2(4.25) + 3(2.75) = 9.25 + 8.5 + 8.25 = 26 $$ ✔️

All equations are satisfied with the obtained values.
Each bundle yields the following units: - High-grade grain: 9.25 units - Medium-grade grain: 4.25 units - Low-grade grain: 2.75 units

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