Difference between revisions of "2003 AMC 10A Problems/Problem 13"
| Line 38: | Line 38: | ||
Or, in matrix form | Or, in matrix form | ||
| − | + | <math> | |
\begin{bmatrix} | \begin{bmatrix} | ||
1 & 1 & 1 \\ | 1 & 1 & 1 \\ | ||
| Line 55: | Line 55: | ||
0 \\ | 0 \\ | ||
\end{bmatrix} | \end{bmatrix} | ||
| − | + | </math> | |
To solve this matrix equation, we can rearrange it thus: | To solve this matrix equation, we can rearrange it thus: | ||
| + | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
x \\ | x \\ | ||
| Line 74: | Line 75: | ||
0 \\ | 0 \\ | ||
\end{bmatrix} | \end{bmatrix} | ||
| − | + | </math> | |
Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields | Solving this matrix equation by using [[inverse matrices]] and [[matrix multiplication]] yields | ||
| + | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
x \\ | x \\ | ||
| Line 87: | Line 89: | ||
16 \\ | 16 \\ | ||
\end{bmatrix} | \end{bmatrix} | ||
| + | </math> | ||
Which means that x = 0.5, y = 3.5, and z = 16. Therefore, xyz = (0.5)(3.5)(16) = 28 | Which means that x = 0.5, y = 3.5, and z = 16. Therefore, xyz = (0.5)(3.5)(16) = 28 | ||
Revision as of 17:22, 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 = 0.5, y = 3.5, and z = 16. Therefore, xyz = (0.5)(3.5)(16) = 28
