Question

The top three teams qualified to compete at the state competition. At the state competition, however, scoring is done differently than at the other meets. Use the following information to create a system of equations where is the points for first place, is the points for second place, and is the points for third place.

Results:

Eagles got 5 first-place, 3 second-place, and 4 third-place finishes with a team total of 67 points.
Pirates got 2 first-place, 5 second-place, and 5 third-place finishes with a team total of 56.
Devil Rays got 6 first-place, 3 second-place and 3 third-place finishes with a team total of 72.
We can’t believe the Devil Rays came from behind and won! We need to determine how many points were awarded for each place. Use Cramer’s rule or inverse method to get the answers. Explain your process.

Answer 1

• first place be x
• second place be y
• third place be z

The equations are

• 5x+3y+4z=67
• 2x+5y+5z=56
• 6x+3y+3z=72

Lets solve them through matrix

Let

`\\ \rm\Rrightarrow A=\left[\begin{array}{ccc}\rm 5&\rm 3&\rm 4\\ \rm 2&\rm 5&\rm 5\\ \rm 6&\rm 3&\rm3\end{array}\right]`

`\\ \rm\Rrightarrow X=\left[\begin{array}{c}\rm x\\ \rm y\\ \rm z\end{array}\right]`

`\\ \rm\Rrightarrow B=\left[\begin{array}{c}\rm 67 \\ \rm 56\\ \rm 72\end{array}\right]`

Let solve

Determinant of A

`\\ \rm\Rrightarrow |A|`

`\\ \rm\Rrightarrow 5\left|\begin{array}{cc}\rm 5&\rm 5\\ \rm 3&\rm 3\end{array}\right|-3\left|\begin{array}{cc}\rm 2&\rm 5\\ \rm 6&\rm 3\end{array}\right|+4\left|\begin{array}{cc}\rm 2&\rm 5\\ \rm 6&\rm 3\end{array}\right|`

`\\ \rm\Rrightarrow 5(15-15)-3(6-30)+4(6-30)`

`\\ \rm\Rrightarrow 72-96`

`\\ \rm\Rrightarrow -24`

Cofactor

`\\ \rm\Rrightarrow A_(11)=0`

`\\ \rm\Rrightarrow A_(12)=24`

`\\ \rm\Rrightarrow A_(13)=-24`

`\\ \rm\Rrightarrow A_(21)=3`

`\\ \rm\Rrightarrow A_(22)=-9`

`\\ \rm\Rrightarrow A_(23)=3`

`\\ \rm\Rrightarrow A_(31)=-5`

`\\ \rm\Rrightarrow A_(32)=-17`

`\\ \rm\Rrightarrow A_(33)=19`

Cofactor of A

`\\ \rm\Rrightarrow \left[\begin{array}{ccc}\rm 0&\rm 24&\rm -24\\ \rm 3&\rm -9&\rm 3\\ \rm -5&\rm -17&\rm 19\end{array}\right]`

Adjacent of A

`\\ \rm\Rrightarrow \left[\begin{array}{ccc}\rm 0&\rm 24&\rm -24\\ \rm 3&\rm -9&\rm 3\\ \rm -5&\rm -17&\rm 19\end{array}\right]^T`

`\\ \rm\Rrightarrow \left[\begin{array}{ccc}\rm 0&\rm 3&\rm -5\\ \rm 24&\rm -9&\rm -17\\ \rm -24&\rm 3&\rm 19\end{array}\right]`

Inverse of A

`\\ \rm\Rrightarrow (1)/(|A|)* Adj.A`

`\\ \rm\Rrightarrow (1)/(-24)\left[\begin{array}{ccc}\rm 0&\rm 3&\rm -5\\ \rm 24&\rm -9&\rm -17\\ \rm -24&\rm 3&\rm 19\end{array}\right]`

Now

`\\ \rm\Rrightarrow AX=B`

`\\ \rm\Rrightarrow X=A^(-1)B`

`\\ \rm\Rrightarrow X=(1)/(-24)\left[\begin{array}{ccc}\rm 0&\rm 3&\rm -5\\ \rm 24&\rm -9&\rm -17\\ \rm -24&\rm 3&\rm 19\end{array}\right]\left[\begin{array}{c}\rm 67\\ \rm 56\\ \rm 72\end{array}\right]`

`\\ \rm\Rrightarrow X=(1)/(-24)\left[\begin{array}{c}\rm -192\\ \rm -120\\ \rm -72\end{array}\right]`

`\\ \rm\Rrightarrow X=\left[\begin{array}{c}\rm 8\\ \rm 5\\ \rm 3\end{array}\right]`

So

`\\ \rm\Rrightarrow \left[\begin{array}{c}\rm x\\ \rm y\\ \rm z\end{array}\right]=\left[\begin{array}{c}\rm 8\\ \rm 5\\ \rm 3\end{array}\right]`

We get

• x=8
• y=5
• z=3