Difference between revisions of "2003 AMC 10A Problems/Problem 13"
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= | = | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | + | \frac{1}{2} \\ | |
| − | + | \frac{7}{2} \\ | |
16 \\ | 16 \\ | ||
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| − | Which means that x = | + | Which means that x = \frac{1}{2}, y = \frac{7}{2}, and z = 16. Therefore, xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28 |
== See Also == | == See Also == | ||
Revision as of 18:27, 21 November 2007
Problem
The sum of three numbers is 
. The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?
Solution
Let the numbers be 
, 
, and 
in that order.
Therefore, the product of all three numbers is 
Alternatively, we can set up the system in matrix form:
is equivalent to
Or, in matrix form
To solve this matrix equation, we can rearrange it thus:
Solving this matrix equation by using inverse matrices and matrix multiplication yields
Which means that x = \frac{1}{2}, y = \frac{7}{2}, and z = 16. Therefore, xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28
