Calculate the covariance matrix for three random variables

Question

I have the task:

The random values $X$, $Y$ and $Z$ are related by the ratio $2X -3Y + Z = 0$. Moreover, $EX = EY = EZ = 0$, $DX = 2, DY = 1, DZ = 5$.

Calculate the covariance matrix $\Lambda$ of these random variables.

I have found the matrix $\Lambda$ and $\det \Lambda = 0$. It is known that the covariance matrix can be singular.

Question Is it possible a direct calculation without liner algebra?

My solution is:

We need find the matrix: \begin{equation}\Lambda=\begin{bmatrix} var(X) & cov(X,Y) & cov(X,Z) \\ cov(Y,X) & var(Y) & cov(Y,Z) \\ cov(Z,X) & cov(Z,Y) & var(Z) \end{bmatrix} \end{equation}

From the statement: $var(X)=DX=2$, $var(Y)=DY=1$, $var(Z)=DZ=5$, $$cov(X,Y)=E(XY)-E(X)E(Y)=E(XY),$$ $$cov(X,Z)=E(XZ)-E(X)E(Z)=E(XZ),$$ and $$cov(Y,Z)=E(YZ)-E(Y)E(Z)=E(YZ).$$ So, now we need compute $E(XY)$, $E(XZ)$, and $E(YZ)$.

1. Take the ratio $2X -3Y + Z = 0$, multiply by $X$, $Y$ and $Z$ and construct the system: \begin{cases} 2\cdot X\cdot X -3\cdot Y \cdot X + Z\cdot X & = & 0 \\ 2\cdot X\cdot Y -3\cdot Y \cdot Y + Z\cdot Y & = & 0 \\ 2\cdot X\cdot Z -3\cdot Y \cdot Z + Z\cdot Z & = & 0 \\ \end{cases}

2. Apply the expectation $E$ function:

\begin{cases} 2\cdot E(X\cdot X) -3\cdot E(Y \cdot X) + E(Z\cdot X) & = & 0 \\ 2\cdot E(X\cdot Y) -3\cdot E(Y \cdot Y) + E(Z\cdot Y) & = & 0 \\ 2\cdot E(X\cdot Z) -3\cdot E(Y \cdot Z) + E(Z\cdot Z) & = & 0 \\ \end{cases}

3. From the statement: $E(XX)=D(X)=2$, $E(YY)=D(Y)=1$, $E(ZZ)=D(Z)=5$, then \begin{cases} 2\cdot 2 -3\cdot E(Y \cdot X) + E(Z\cdot X) & = & 0 \\ 2\cdot E(X\cdot Y) -3\cdot 1 + E(Z\cdot Y) & = & 0 \\ 2\cdot E(X\cdot Z) -3\cdot E(Y \cdot Z) + 5 & = & 0 \\ \end{cases} or \begin{bmatrix} -3 & 1 & 0 &|&-4\\ 2 & 0 & 1 &|&3\\ 0 & 2 & -3 &|&-5 \end{bmatrix}

The solutuon is: $E(XY)=1$, $E(XZ)=-1$, and $E(YZ)=1$.

4. Finally \begin{equation}\Lambda=\begin{bmatrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ -1 & 1 & 5 \end{bmatrix} \end{equation} and $\det \Lambda = 0$.