Let $X_1,...,X_n$ be random variables with covariance matrix $\Lambda_1$, and let $Y_1,...,Y_{n-1}$ be random variables with covariance matrix $\Lambda_2$. Using $\Lambda_1$ and $\Lambda_2$, denote the following matrix:
$\Lambda = \begin{pmatrix}
\Lambda_2 & a \\
a^t & Var(X_n)
\end{pmatrix}$
Where
$a = \begin{pmatrix}
cov(X_1,X_n), &..., & cov(X_{n-1},X_n)
\end{pmatrix}^t$
Can we say $\Lambda$ is also a covariance matrix?
$\Lambda$ can either be a covariance matrix or not be a covariance matrix. It depends on $(X_1,..,X_n)$ and $(Y_1,..,Y_{n-1})$.
If these $(2n-1)$ random variables are independent to each other, then $\Lambda$ is a diagonal matrix and so, it's a covariance matrix.
If for example, $n = 3$ and $X_1 =X_2=X_3$, and $(Y_1, Y_2)$ are independent to each other. suppose their variances are all equal to $1$, then $$\Lambda=\pmatrix{1&0&1\\0&1&1\\1&1&1}$$ A covariance matrix need to be positive definite. However, the matrix $\Lambda$ is not positive definite as its eigenvalues are $(1,1+\sqrt{2},1-\sqrt{2})$ (the third eigenvalue is negative. To be a positive defenite matrix, all eigenvalues need to be positive).