Question

During the 2009-2010 season, inter milan of the Italian series A scored 82 points on 38 rounds. For each round of the game, the winning team would receive 3 points, while the losing team would get 0 points. But if the game ends up in a tie, each team would get 1 point. Inter Milan won 20 more rounds than they lost. Use a matrix to find how many of each type of shot they made

Answer 1

Inter milan had 24 wins, 10 draws and 4 losses.

Explanation:

Let n₁ = number of wins, n₂ = number of draws and n₃ = number of number of losses. Let W = win points = 3, L = loss points = 0 and D = draw points = 1 Since we have our total number of points a 82, then

n₁W + n₂D + n₃L = 82

3n₁ + n₂ + 0n₃ = 82 (1)

Also, since there are 38 points in total, we have that

n₁ + n₂ + n₃ = 38 (2)

Also, we have 20 more win points than loss points. So,

n₁ = n₃ + 20

n₁ + 0n₂ - n₃ = 20 (3)

We have 3 equations. We now write them in matrix form below.

`\left[\begin{array}{ccc}3&1&0\\1&1&1\\1&0&-1\end{array}\right] \left[\begin{array}{ccc}n_(1) \\n_(2)\\n_(3)\end{array}\right] = \left[\begin{array}{ccc}82 \\38\\20\end{array}\right]`

In the form AX = B, where A =

`A =\left[\begin{array}{ccc}3&1&0\\1&1&1\\1&0&-1\end{array}\right] , X =\left[\begin{array}{ccc}n_(1) \\n_(2)\\n_(3)\end{array}\right] and B= \left[\begin{array}{ccc}82 \\38\\20\end{array}\right]`

Using cramer's rule, n₁ =

`det\left[\begin{array}{ccc}82&1&0\\38&1&1\\20&0&-1\end{array}\right] /det\left[\begin{array}{ccc}3&1&0\\1&1&1\\1&0&-1\end{array}\right]`

= 82((-1 × 1) - 0 × 1) - 1((38 × -1)- (20 × 1)) + 0((38 × 0) - 20 × 1) ÷ 3((-1 × 1) - 0 × 1) - 1((-1 × 1) - 1 × 1) + 0(1 × 0 + 1 × 1)

= (-82 + 58 + 0)/(-3 + 2 + 0)

= -24/-1

= 24

n₂ =

`det\left[\begin{array}{ccc}3&82&0\\1&38&1\\1&20&-1\end{array}\right] /det\left[\begin{array}{ccc}3&1&0\\1&1&1\\1&0&-1\end{array}\right]`

= 3((38 × -1)- (20 × -1)) - 82((-1 × 1) - 1 × 1) + 0(20 × 1 - (38 × 1) ) ÷ 3((-1 × 1) - 0 × 1) - 1((-1 × 1) - 1 × 1) + 0(1 × 0 + 1 × 1)

= (-174 + 164 + 0)/(-3 + 2 + 0)

= -10/-1

= 10

n₃ =

`det\left[\begin{array}{ccc}3&1&82\\1&1&38\\1&0&20\end{array}\right] /det\left[\begin{array}{ccc}3&1&0\\1&1&1\\1&0&-1\end{array}\right]`

= 3((1 × 20)- (0 × 38)) - 1((1 × 20) - 1 × 38) + 82(1 × 0 - (1 × 1) ) ÷ 3((-1 × 1) - 0 × 1) - 1((-1 × 1) - 1 × 1) + 0(1 × 0 + 1 × 1)

= (60 + 18 - 82)/(-3 + 2 + 0)

= -4/-1

= 4

So, Inter milan had 24 wins, 10 draws and 4 losses.