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Q67E

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Found in: Page 23

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Five cows and two sheep together cost 10 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)

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

See the step by step solution

## Step 1: 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.

$5C+3S=10\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)$

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

$2C+5S=8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)$

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

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

Represent the above equations in terms of a matrix.

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

## Step 2: Perform the row operations on the matrix

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

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

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.

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