Question

Five cows and two sheep together cost10 liang of silver. Two cows and five sheep together cost eight liang of silver. What is the cost of a cow and a sheep, respectively? (Nine Chapters, Chapter 8, Problem 7)

Short answer

$\frac{34}{21}$liang is the cost of a cow and$\frac{20}{21}$ liang is the cost of a sheep.

Step-by-step solution

Represent the cost of cows and sheep in terms of a system of equations.  

The cost of a cow and a sheep can be found using the augmented matrix.

$5$cows and$2$sheep together cost$10$liang of silver.

$5C+3S=10\mathrm{......}\left(1\right)$

$2$cows and$5$sheep together cost$8$liang of silver.

$2C+5S=8\mathrm{......}\left(2\right)$

Consider the equations (1) and (2). The system of equations is,

$\left|\begin{array}{l}5C+3S=10\\ 2C+5S=8\end{array}\right|$

Represent the above equations in terms of a matrix.

$\left|\begin{array}{ccc}5& 3& 10\\ 2& 5& 8\end{array}\right|$

Perform the row operations on the matrix

Consider the obtained matrix, now reduce it into row echelon form.

$\mathbf{rref}\mathbf{\left|}\begin{array}{ccc}\mathbf{5}& \mathbf{3}& \mathbf{10}\\ \mathbf{2}& \mathbf{5}& \mathbf{8}\end{array}\mathbf{\right|}\mathbf{\to }\mathbf{\left|}\begin{array}{ccc}\mathbf{1}& \mathbf{0}& \frac{\mathbf{34}}{\mathbf{21}}\\ \mathbf{0}& \mathbf{1}& \frac{\mathbf{20}}{\mathbf{21}}\end{array}\mathbf{\right|}$

The values are, $C=\frac{34}{21},S=\frac{20}{21}$.

Hence, $\frac{34}{21}$liang is the cost of a cow and $\frac{20}{21}$liang is the cost of a sheep.