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 art” 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 1 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?

Question

UpStudy AI · Accepted Answer

Certainly! Let's solve the given system of equations using matrix methods. The problem involves finding the unit values of high-grade ($$x$$), medium-grade ($$y$$), and low-grade ($$z$$) grain bundles based on the provided equations.

### **Given System of Equations:**
1. $$3x + 2y + z = 39$$
2. $$2x + 3y + z = 34$$
3. $$x + 2y + 3z = 26$$

### **Step 1: Represent the System as an Augmented Matrix**
$$
\begin{bmatrix}
3 & 2 & 1 & | & 39 \
2 & 3 & 1 & | & 34 \
1 & 2 & 3 & | & 26 \
\end{bmatrix}
$$

### **Step 2: Perform Row Operations to Achieve Row-Echelon Form**

**a. Eliminate $$x$$ from the second and third rows:**

- **R2' = R2 - (2/3)R1**
$$
R2' = \begin{bmatrix} 2 & 3 & 1 & | & 34 \end{bmatrix} - \frac{2}{3} \begin{bmatrix} 3 & 2 & 1 & | & 39 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{3} & \frac{1}{3} & | & 34 - 26 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{3} & \frac{1}{3} & | & 8 \end{bmatrix}
$$

- **R3' = R3 - (1/3)R1**
$$
R3' = \begin{bmatrix} 1 & 2 & 3 & | & 26 \end{bmatrix} - \frac{1}{3} \begin{bmatrix} 3 & 2 & 1 & | & 39 \end{bmatrix} = \begin{bmatrix} 0 & \frac{4}{3} & \frac{8}{3} & | & 26 - 13 \end{bmatrix} = \begin{bmatrix} 0 & \frac{4}{3} & \frac{8}{3} & | & 13 \end{bmatrix}
$$

**b. Update the Matrix:**
$$
\begin{bmatrix}
3 & 2 & 1 & | & 39 \
0 & \frac{5}{3} & \frac{1}{3} & | & 8 \
0 & \frac{4}{3} & \frac{8}{3} & | & 13 \
\end{bmatrix}
$$

**c. Eliminate $$y$$ from the third row:**

- Multiply R2' by 4 and R3' by 5 to make the coefficients of $$y$$ the same:
$$
R2'' = 4 \times R2' = \begin{bmatrix} 0 & \frac{20}{3} & \frac{4}{3} & | & 32 \end{bmatrix}
$$
$$
R3'' = 5 \times R3' = \begin{bmatrix} 0 & \frac{20}{3} & \frac{40}{3} & | & 65 \end{bmatrix}
$$

- **R3''' = R3'' - R2''**
$$
R3''' = \begin{bmatrix} 0 & 20/3 & 40/3 & | & 65 \end{bmatrix} - \begin{bmatrix} 0 & 20/3 & 4/3 & | & 32 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 36/3 & | & 33 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 12 & | & 33 \end{bmatrix}
$$

**d. Update the Matrix:**
$$
\begin{bmatrix}
3 & 2 & 1 & | & 39 \
0 & \frac{5}{3} & \frac{1}{3} & | & 8 \
0 & 0 & 12 & | & 33 \
\end{bmatrix}
$$

### **Step 3: Back Substitution**

**a. From the third row:**
$$
12z = 33 \implies z = \frac{33}{12} = \frac{11}{4} = 2.75
$$

**b. Substitute $$z$$ into the second row:**
$$
\frac{5}{3}y + \frac{1}{3} \times \frac{11}{4} = 8 \
\frac{5}{3}y + \frac{11}{12} = 8 \
\frac{5}{3}y = 8 - \frac{11}{12} \
\frac{5}{3}y = \frac{96}{12} - \frac{11}{12} = \frac{85}{12} \
y = \frac{85}{12} \times \frac{3}{5} = \frac{85}{20} = \frac{17}{4} = 4.25
$$

**c. Substitute $$y$$ and $$z$$ into the first row:**
$$
3x + 2 \times \frac{17}{4} + \frac{11}{4} = 39 \
3x + \frac{34}{4} + \frac{11}{4} = 39 \
3x + \frac{45}{4} = 39 \
3x = 39 - \frac{45}{4} = \frac{156}{4} - \frac{45}{4} = \frac{111}{4} \
x = \frac{111}{12} = \frac{37}{4} = 9.25
$$

### **Final Solution:**
$$
\begin{cases}
x = \frac{37}{4} \text{ units} \
y = \frac{17}{4} \text{ units} \
z = \frac{11}{4} \text{ units} \
\end{cases}
$$

Or in mixed numbers:
$$
\begin{cases}
x = 9 \frac{1}{4} \text{ units} \
y = 4 \frac{1}{4} \text{ units} \
z = 2 \frac{3}{4} \text{ units} \
\end{cases}
$$

### **Verification:**

Let's verify by substituting the values back into the original equations:

1. $$3(9.25) + 2(4.25) + 2.75 = 27.75 + 8.5 + 2.75 = 39$$ ✔️
2. $$2(9.25) + 3(4.25) + 2.75 = 18.5 + 12.75 + 2.75 = 34$$ ✔️
3. $$9.25 + 2(4.25) + 3(2.75) = 9.25 + 8.5 + 8.25 = 26$$ ✔️

All equations are satisfied.

**Therefore, the unit values are:**
- **High-grade grain bundle:** $$\frac{37}{4}$$ units (9¼ units)
- **Medium-grade grain bundle:** $$\frac{17}{4}$$ units (4¼ units)
- **Low-grade grain bundle:** $$\frac{11}{4}$$ units (2¾ units)
Each bundle yields the following units: - High-grade grain: 9.25 units - Medium-grade grain: 4.25 units - Low-grade grain: 2.75 units

Upstudy AI Solution

Answer

Solution

Given System of Equations:

Step 1: Represent the System as an Augmented Matrix

Step 2: Perform Row Operations to Achieve Row-Echelon Form

Step 3: Back Substitution

Final Solution:

Verification:

Mind Expander

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