IMC 2019 · Day 1 · P2

Hint: If the solution of the system is not unique then $\det \begin{pmatrix} Y & E & A & R \\ R & Y & E & A \\ A & R & Y & E \\ E & A & R & Y \end{pmatrix} = 0$.

Let us apply row transformations to the augmented matrix of the system to find its rank. First we add the second, third and fourth row to the first one and divide by $Y + E + A + R$ to get $$\left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 1 \\ R & Y & E & A & E \\ A & R & Y & E & A \\ E & A & R & Y & R \end{array} \right) \sim \left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 1 \\ 0 & Y-R & E-R & A-R & E-R \\ 0 & R-A & Y-A & E-A & 0 \\ 0 & A-E & R-E & Y-E & R-E \end{array} \right) \sim \left( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 1 \\ 0 & Y-R & E-R & A-R & E-R \\ 0 & R-A & Y-A & E-A & 0 \\ 0 & A-E+Y-R & 0 & Y-E+A-R & 0 \end{array} \right)$$ Let us first omit the last column and look at the remaining $4 \times 4$ matrix. If $E \ne R$, the first and second rows are linearly independent, so the rank of the matrix is at least 2 and rank of the augmented $4 \times 5$ matrix cannot be bigger than rank of the $4 \times 4$ matrix due to the zeros in the last column.

IF $E = R$, then we have three zeros in the last column, so rank of the $4 \times 5$ matrix cannot be bigger than rank of the $4 \times 4$ matrix. So, the original system has always at least one solution.

It follows that the system has more than one solution if and only if the $4 \times 4$ matrix (with the last column omitted) is singular. Let us first assume that $E \ne R$. We apply one more transform to get $$\sim \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & Y-R & E-R & A-R \\ 0 & (R-A)(E-R) - (Y-R)(Y-A) & 0 & (E-A)(E-R) - (A-R)(Y-A) \\ 0 & A-E+Y-R & 0 & Y-E+A-R \end{pmatrix}$$ Obviously, this matrix is singular if and only if $A - E + Y - R = 0$ or the two expressions in the third row are equal, i.e. $$\begin{gather*} RE - R^2 - AE + AR - Y^2 + RY + AY - AR = E^2 - AE - ER + AR - AY + RY + A^2 - AR \\ 0 = (E - R)^2 + (A - Y)^2, \end{gather*}$$ but this is impossible if $E \ne R$. If $E = R$, we have $$\sim \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & Y-R & 0 & A-R \\ 0 & R-A & Y-A & R-A \\ 0 & A+Y-2R & 0 & Y+A-2R \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & Y-R & 0 & A-R \\ 0 & R-A & Y-A & R-A \\ 0 & A-R & 0 & Y-R \end{pmatrix}.$$ If $A = Y$, this matrix is singular. If $A \ne Y$, the matrix is regular if and only if $(Y - R)^2 \ne (A - R)^2$ and since $Y \ne A$, it means that $Y - R \ne -(A - R)$, i.e. $Y + A \ne 2R$. We conclude that $YEAR$ is very good if and only if

1. $E \ne R$ and $A + Y = E + R$, or

2. $E = R$ and $Y = A$, or

3. $E = R$, $A \ne Y$ and $Y + A = 2R$.

We can see that if $Y = 2$, $E = 0$, then the very good years satisfying 1 are $A + 2 = R \ne 0$, i.e. 2002, 2013, 2024, 2035, 2046, 2057, 2068, 2079, condition 2 is satisfied for 2020 and condition 3 never satisfied.