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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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