An efficient cross-check uses the invariants of the matrix instead of expanding the full determinant symbolically. The trace of $A$ (sum of the diagonal entries) is $2+5+10=17$, and this must equal the sum of the eigenvalues. Only options where the three eigenvalues add up to $17$ can be correct. Checking option (C), $1 + (8+\sqrt{37}) + (8-\sqrt{37}) = 1+16 = 17$, which matches. Checking option (A), $1+(8-\sqrt{37})+(-8+\sqrt{37}) = 1+0 = 1 \ne 17$, so that option is ruled out immediately. Option (B) gives $2 + 0 = 2 \ne 17$, also ruled out. Option (D) gives $2+2(8+\sqrt{37}) = 18+2\sqrt{37} \approx 30.2 \ne 17$, ruled out as well. As a second check, the determinant of $A$ must equal the product of the eigenvalues; expanding $\det(A)$ directly gives $\det(A) = 2(50-24)-2(20-18)+3(8-15) = 52-4-21=27$. For option (C), the product is $1\times(8+\sqrt{37})(8-\sqrt{37}) = 1\times(64-37)=27$, which matches perfectly. Both invariants confirm the same set.
\[\boxed{\lambda = 1,\ 8+\sqrt{37},\ 8-\sqrt{37}}\]For the matrix, $A = \begin{bmatrix} -4 & 0 \\ -1.6 & 4 \end{bmatrix}$, the eigenvalues ($\lambda$) and eigenvectors ($X$) respectively are:
Consider the following matrix: \[ \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix} \] The largest eigenvalue of the above matrix is \(\underline{\hspace{2cm}}\).