Question

If \(\lambda_1,\lambda_2,\lambda_3\) are the eigenvalues of the matrix \[\begin{pmatrix}-2&2&-3\\2&1&-6\\-1&-2&0\end{pmatrix}\] then the value of \(\lambda_1^2+\lambda_2^2+\lambda_3^2\) is ____.

• A. 40
• B. 43
• C. 45
• D. 48

Hint:

Use trace(A^2) = sum of lambda_i^2, computed directly from the diagonal entries of A squared.

Answer 1

The Correct Option is B

Instead of using trace identities, find the eigenvalues explicitly and square them.

The characteristic polynomial of \(A\) is \(\det(A-\lambda I)=0\). Using \(\operatorname{trace}(A) = -1\), the determinant \(\det(A) = 45\) (computed by cofactor expansion), and the sum of principal \(2\times2\) minors equal to \(-21\), the characteristic equation is \[-\lambda^3 + (\operatorname{trace} A)\lambda^2 - (\text{sum of minors})\lambda + \det(A) = 0\] which rearranges to \[\lambda^3 + \lambda^2 - 21\lambda - 45 = 0\]

Test small integer divisors of 45 as candidate roots. Trying \(\lambda = 5\): \[5^3+5^2-21(5)-45 = 125+25-105-45 = 0\] So \(\lambda=5\) is a root. Dividing \(\lambda^3+\lambda^2-21\lambda-45\) by \((\lambda-5)\) gives \[\lambda^2+6\lambda+9 = (\lambda+3)^2\]

So the full factorization is \((\lambda-5)(\lambda+3)^2=0\), giving eigenvalues \[\lambda_1 = 5, \qquad \lambda_2=-3, \qquad \lambda_3=-3\]

Now square and add them directly: \[\lambda_1^2+\lambda_2^2+\lambda_3^2 = 5^2+(-3)^2+(-3)^2 = 25+9+9 = 43\] This agrees exactly with the trace-based computation.

\[\boxed{\lambda_1^2+\lambda_2^2+\lambda_3^2 = 43}\]