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I am going to set up a system of linear equations. Of course, this means I have to solve for what that system is before I can solve the system by algebra, and therein line the trick.

I will use the notation $x_{ij}$ for the goals scored by Team $i$ against Team $j$. The numbers will be the rankings of the teams, thus Spain = 1, Italy = 2, Croatia = 3 and Albania = 4.

Spain was massive on defense so we force $x_{21}=x_{31}=x_{41}=0$. We drop those from the system and now seek nine equations for the other nine variables.

From the goals for and goals against values given in the rable we may render Eqs. (1)-(7):

$x_{12}+x_{13}+x_{14}=5\tag{1}$ $x_{23}+x_{24}=3\tag{2}$ $x_{12}+x_{32}+x_{42}=3\tag{3}$

We do not assume that each the terms in Eq. (3) is one goal because Italy won a game (can't otherwise have four points) and may have pitched a shutout in that game.

$x_{32}+x_{34}=3\tag{4}$ $x_{13}+x_{23}+x_{43}=6\tag{5}$ $x_{42}+x_{43}=3\tag{6}$ $x_{14}+x_{24}+x_{34}=5\tag{7}$

Now Croatia gained two points and must have lost to Spain, so it drew against the other two teams and we now have, ostensibly, the two missing equations: $x_{23}-x_{32}=0\tag{8}$ $x_{24}-x_{42}=0\tag{9}$

We have our nine equations for the nine unknowns and are ready to go ... until we check the determinant of the coefficient matrix: $0$! Our intended nine equations fail to be linearly independent!

Suppose we label the coefficient matrix from the above equation $M$, the identity matrix $I$, and a small scalar $\epsilon$. Plugging in various values of $\epsilon$ we find that with $\epsilon$ indeed small in absolute value, the determinant of $M-\epsilon I$ tends to proportionality with $\epsilon^2$. This indicates two zero eigenvalues, thus $9-2=7$ of the above nine equations are actually linearly independent. Using principal minors we find that one such set of seven equations is (2) through (8). We therefore drop Eqs. (1) and (9) and must replace them with equations that are independent of the remaining seven.

Albania must have suffered some heartbreakers. They never won but their goal differential was merely -2, so they must have taken two losses by one goal plus their draw. With Spain shutting out everyone else, this forces $x_{14}=1\tag{10}$

Now we look at Albania versus Italy. If that were a draw, then Italy would have had to win over Spain (but Spain won all three of its matches) or against Croatia (which was proved earlier to have drawn except against Spain). The contradiction forces us to accept that Albania lost to Italy, by one goal (and drew against Croatia): $x_{24}-x_{42}=1\tag{11}$

We now have Eqs. (2) through (8) and now also (10) and (11). This time the coefficient matrix has a nonzero determinant and the system with those nine equations can be solved. The solution, with implied scores, is as follows (pardon the multiple paragraphs; there were formatting issues in my iriginal display):

$x_{12}=1\implies$ Spain 1, Italy 0

$x_{13}=3\implies$ Spain 3, Croatia 0

$x_{14}=1\implies$ Spain 1, Albania 0

$x_{23}=1,x_{32}=1\implies$ Italy 1, Croatia 1

$x_{24}=2,x_{42}=1\implies$ Italy 2, Albania 1

$x_{34}=2,x_{43}=2\implies$ Croatia 2, Albania 2

I am going to set up a system of linear equations. Of course, this means I have to solve for what that system is before I can solve the system by algebra, and therein line the trick.

I will use the notation $x_{ij}$ for the goals scored by Team $i$ against Team $j$. The numbers will be the rankings of the teams, thus Spain = 1, Italy = 2, Croatia = 3 and Albania = 4.

Spain was massive on defense so we force $x_{21}=x_{31}=x_{41}=0$. We drop those from the system and now seek nine equations for the other nine variables.

From the goals for and goals against values given in the rable we may render Eqs. (1)-(7):

$x_{12}+x_{13}+x_{14}=5\tag{1}$ $x_{23}+x_{24}=3\tag{2}$ $x_{12}+x_{32}+x_{42}=3\tag{3}$

We do not assume that each the terms in Eq. (3) is one goal because Italy won a game (can't otherwise have four points) and may have pitched a shutout in that game.

$x_{32}+x_{34}=3\tag{4}$ $x_{13}+x_{23}+x_{43}=6\tag{5}$ $x_{42}+x_{43}=3\tag{6}$ $x_{14}+x_{24}+x_{34}=5\tag{7}$

Now Croatia gained two points and must have lost to Spain, so it drew against the other two teams and we now have, ostensibly, the two missing equations: $x_{23}-x_{32}=0\tag{8}$ $x_{24}-x_{42}=0\tag{9}$

We have our nine equations for the nine unknowns and are ready to go ... until we check the determinant of the coefficient matrix: $0$! Our intended nine equations fail to be linearly independent!

Suppose we label the coefficient matrix from the above equation $M$, the identity matrix $I$, and a small scalar $\epsilon$. Plugging in various values of $\epsilon$ we find that with $\epsilon$ indeed small in absolute value, the determinant of $M-\epsilon I$ tends to proportionality with $\epsilon^2$. This indicates two zero eigenvalues, thus $9-2=7$ of the above nine equations are actually linearly independent. Using principal minors we find that one such set of seven equations is (2) through (8). We therefore drop Eqs. (1) and (9) and must replace them with equations that are independent of the remaining seven.

Albania must have suffered some heartbreakers. They never won but their goal differential was merely -2, so they must have taken two losses by one goal plus their draw. With Spain shutting out everyone else, this forces $x_{14}=1\tag{10}$

Now we look at Albania versus Italy. If that were a draw, then Italy would have had to win over Spain (but Spain won all three of its matches) or against Croatia (which was proved earlier to have drawn except against Spain). The contradiction forces us to accept that Albania lost to Italy, by one goal (and drew against Croatia): $x_{24}-x_{42}=1\tag{11}$

We now have Eqs. (2) through (8) and now also (10) and (11). This time the coefficient matrix has a nonzero determinant and the system with those nine equations can be solved. The solution, with implied scores, is as follows (pardon the multiple paragraphs; there were formatting issues in my original display):

$x_{12}=1\implies$ Spain 1, Italy 0

$x_{13}=3\implies$ Spain 3, Croatia 0

$x_{14}=1\implies$ Spain 1, Albania 0

$x_{23}=1,x_{32}=1\implies$ Italy 1, Croatia 1

$x_{24}=2,x_{42}=1\implies$ Italy 2, Albania 1

$x_{34}=2,x_{43}=2\implies$ Croatia 2, Albania 2

I am going to set up a system of linear equations. Of course, this means I have to solve for what that system is before I can solve the system by algebra, and therein line the trick.

I will use the notation $x_{ij}$ for the goals scored by Team $i$ against Team $j$. The numbers will be the rankings of the teams, thus Spain = 1, Italy = 2, Croatia = 3 and Albania = 4.

Spain was massive on defense so we force $x_{21}=x_{31}=x_{41}=0$. We drop those from the system and now seek nine equations for the other nine variables.

From the goals for and goals against values given in the rable we may render Eqs. (1)-(7):

$x_{12}+x_{13}+x_{14}=5\tag{1}$ $x_{23}+x_{24}=3\tag{2}$ $x_{12}+x_{32}+x_{42}=3\tag{3}$

We do not assume that each the terms in Eq. (3) is one goal because Italy won a game (can't otherwise have four points) and may have pitched a shutout in that game.

$x_{32}+x_{34}=3\tag{4}$ $x_{13}+x_{23}+x_{43}=6\tag{5}$ $x_{42}+x_{43}=3\tag{6}$ $x_{14}+x_{24}+x_{34}=5\tag{7}$

Now Croatia gained two points and must have lost to Spain, so it drew against the other two teams and we now have, ostensibly, the two missing equations: $x_{23}-x_{32}=0\tag{8}$ $x_{24}-x_{42}=0\tag{9}$

We have our nine equations for the nine unknowns and are ready to go ... until we check the determinant of the coefficient matrix: $0$! Our intended nine equations fail to be linearly independent!

Suppose we label the coefficient matrix from the above equation $M$, the identity matrix $I$, and a small scalar $\epsilon$. Plugging in various values of $\epsilon$ we find that with $\epsilon$ indeed small in absolute value, the determinant of $M-\epsilon I$ tends to proportionality with $\epsilon^2$. This indicates two zero eigenvalues, thus $9-2=7$ of the above nine equations are actually linearly independent. Using principal minors we find that one such set of seven equations is (2) through (8). We therefore drop Eqs. (1) and (9) and must replace them with equations that are independent of the remaining seven.

Albania must have suffered some heartbreakers. They never won but their goal differential was merely -2, so they must have taken two losses by one goal plus their draw. With Spain shutting out everyone else, this forces $x_{14}=1\tag{10}$

Now we look at Albania versus Italy. If that were a draw, then Italy would have had to win over Spain (but Spain won all three of its matches) or against Croatia (which was proved earlier to have drawn except against Spain). The contradiction forces us to accept that Albania lost to Italy, by one goal (and drew against Croatia): $x_{24}-x_{42}=1\tag{11}$

We now have Eqs. (2) through (8) and now also (10) and (11). This time the coefficient matrix has a nonzero determinant and the system with those nine equations can be solved. The solution, with implied scores, is as follows:

$x_{12}=1\implies$ Spain 1, Italy 0  $x_{13}=3\implies$ Spain 3, Croatia 0  $x_{14}=1\implies$ Spain 1, Albania 0  $x_{23}=1,x_{32}=1\implies$ Italy 1, Croatia 1  $x_{24}=2,x_{42}=1\implies$ Italy 2, Albania 1  $x_{34}=2,x_{43}=2\implies$ Croatia 2, Albania 2

I am going to set up a system of linear equations. Of course, this means I have to solve for what that system is before I can solve the system by algebra, and therein line the trick.

I will use the notation $x_{ij}$ for the goals scored by Team $i$ against Team $j$. The numbers will be the rankings of the teams, thus Spain = 1, Italy = 2, Croatia = 3 and Albania = 4.

Spain was massive on defense so we force $x_{21}=x_{31}=x_{41}=0$. We drop those from the system and now seek nine equations for the other nine variables.

From the goals for and goals against values given in the rable we may render Eqs. (1)-(7):

$x_{12}+x_{13}+x_{14}=5\tag{1}$ $x_{23}+x_{24}=3\tag{2}$ $x_{12}+x_{32}+x_{42}=3\tag{3}$

We do not assume that each the terms in Eq. (3) is one goal because Italy won a game (can't otherwise have four points) and may have pitched a shutout in that game.

$x_{32}+x_{34}=3\tag{4}$ $x_{13}+x_{23}+x_{43}=6\tag{5}$ $x_{42}+x_{43}=3\tag{6}$ $x_{14}+x_{24}+x_{34}=5\tag{7}$

Now Croatia gained two points and must have lost to Spain, so it drew against the other two teams and we now have, ostensibly, the two missing equations: $x_{23}-x_{32}=0\tag{8}$ $x_{24}-x_{42}=0\tag{9}$

We have our nine equations for the nine unknowns and are ready to go ... until we check the determinant of the coefficient matrix: $0$! Our intended nine equations fail to be linearly independent!

Suppose we label the coefficient matrix from the above equation $M$, the identity matrix $I$, and a small scalar $\epsilon$. Plugging in various values of $\epsilon$ we find that with $\epsilon$ indeed small in absolute value, the determinant of $M-\epsilon I$ tends to proportionality with $\epsilon^2$. This indicates two zero eigenvalues, thus $9-2=7$ of the above nine equations are actually linearly independent. Using principal minors we find that one such set of seven equations is (2) through (8). We therefore drop Eqs. (1) and (9) and must replace them with equations that are independent of the remaining seven.

Albania must have suffered some heartbreakers. They never won but their goal differential was merely -2, so they must have taken two losses by one goal plus their draw. With Spain shutting out everyone else, this forces $x_{14}=1\tag{10}$

Now we look at Albania versus Italy. If that were a draw, then Italy would have had to win over Spain (but Spain won all three of its matches) or against Croatia (which was proved earlier to have drawn except against Spain). The contradiction forces us to accept that Albania lost to Italy, by one goal (and drew against Croatia): $x_{24}-x_{42}=1\tag{11}$

We now have Eqs. (2) through (8) and now also (10) and (11). This time the coefficient matrix has a nonzero determinant and the system with those nine equations can be solved. The solution, with implied scores, is as follows (pardon the multiple paragraphs; there were formatting issues in my iriginal display):

$x_{12}=1\implies$ Spain 1, Italy 0 

$x_{13}=3\implies$ Spain 3, Croatia 0 

$x_{14}=1\implies$ Spain 1, Albania 0 

$x_{23}=1,x_{32}=1\implies$ Italy 1, Croatia 1 

$x_{24}=2,x_{42}=1\implies$ Italy 2, Albania 1 

$x_{34}=2,x_{43}=2\implies$ Croatia 2, Albania 2

Spain was massive on defense so we force $x_{21}=x_{31}=x_{41}=0$. We drop those from the system and now seek nine equations for the other nine variables.

From the goals for and goals against values given in the rable we may render Eqs. (1)-(7):

$x_{12}+x_{13}+x_{14}=5\tag{1}$

$x_{23}+x_{24}=3\tag{2}$

$x_{12}+x_{32}+x_{42}=3\tag{3}$

We do not assume that each the terms in Eq. (3) is one goal because Italy won a game (can't otherwise have four points) and may have pitched a shutout in that game.

$x_{32}+x_{34}=3\tag{4}$

$x_{13}+x_{23}+x_{43}=6\tag{5}$

$x_{42}+x_{43}=3\tag{6}$

$x_{14}+x_{24}+x_{34}=5\tag{7}$

Now Croatia gained two points and must have lost to Spain, so it drew against the other two teams and we now have, ostensibly, the two missing equations:

$x_{23}-x_{32}=0\tag{8}$

$x_{24}-x_{42}=0\tag{9}$

Albania must have suffered some heartbreakers. They never won but their goal differential was merely -2, so they must have taken two losses by one goal plus their draw. With Spain shutting out everyone else, this forces

$x_{14}=1\tag{10}$

Now we look at Albania versus Italy. If that were a draw, then Italy would have had to win over Spain (but Spain won all three of its matches) or against Croatia (which was proved earlier to have drawn except against Spain). The contradiction forces us to accept that Albania lost to Italy, by one goal (and drew against Croatia):

$x_{24}-x_{42}=1\tag{11}$

$x_{12}=1\implies$ Spain 1, Italy 0

$x_{13}=3\implies$ Spain 3, Croatia 0

$x_{14}=1\implies$ Spain 1, Albania 0

$x_{23}=1,x_{32}=1\implies$ Italy 1, Croatia 1

$x_{24}=2,x_{42}=1\implies$ Italy 2, Albania 1

$x_{34}=2,x_{43}=2\implies$ Croatia 2, Albania 2

Spain was massive on defense so we force $x_{21}=x_{31}=x_{41}=0$. We drop those from the system and now seek nine equations for the other nine variables.

From the goals for and goals against values given in the rable we may render Eqs. (1)-(7):

$x_{12}+x_{13}+x_{14}=5\tag{1}$ $x_{23}+x_{24}=3\tag{2}$ $x_{12}+x_{32}+x_{42}=3\tag{3}$

We do not assume that each the terms in Eq. (3) is one goal because Italy won a game (can't otherwise have four points) and may have pitched a shutout in that game.

$x_{32}+x_{34}=3\tag{4}$ $x_{13}+x_{23}+x_{43}=6\tag{5}$ $x_{42}+x_{43}=3\tag{6}$ $x_{14}+x_{24}+x_{34}=5\tag{7}$

Now Croatia gained two points and must have lost to Spain, so it drew against the other two teams and we now have, ostensibly, the two missing equations: $x_{23}-x_{32}=0\tag{8}$ $x_{24}-x_{42}=0\tag{9}$

Albania must have suffered some heartbreakers. They never won but their goal differential was merely -2, so they must have taken two losses by one goal plus their draw. With Spain shutting out everyone else, this forces $x_{14}=1\tag{10}$

Now we look at Albania versus Italy. If that were a draw, then Italy would have had to win over Spain (but Spain won all three of its matches) or against Croatia (which was proved earlier to have drawn except against Spain). The contradiction forces us to accept that Albania lost to Italy, by one goal (and drew against Croatia): $x_{24}-x_{42}=1\tag{11}$

$x_{12}=1\implies$ Spain 1, Italy 0 $x_{13}=3\implies$ Spain 3, Croatia 0 $x_{14}=1\implies$ Spain 1, Albania 0 $x_{23}=1,x_{32}=1\implies$ Italy 1, Croatia 1 $x_{24}=2,x_{42}=1\implies$ Italy 2, Albania 1 $x_{34}=2,x_{43}=2\implies$ Croatia 2, Albania 2