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

Step 1: Recall two standard identities relating eigenvalues to the trace of powers of the matrix. If \(A\) has eigenvalues \(\lambda_1,\lambda_2,\lambda_3\), then \[\operatorname{trace}(A) = \lambda_1+\lambda_2+\lambda_3, \qquad \operatorname{trace}(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2\] So the quantity asked for is simply \(\operatorname{trace}(A^2)\).

Step 2: Compute \(A^2\) and read off its diagonal entries. With \[A = \begin{pmatrix}-2&2&-3\\2&1&-6\\-1&-2&0\end{pmatrix}\] the diagonal entries of \(A^2\) come from dotting each row of \(A\) with the corresponding column of \(A\): \[(A^2)_{11} = (-2)(-2)+2(2)+(-3)(-1) = 4+4+3 = 11\] \[(A^2)_{22} = 2(2)+1(1)+(-6)(-2) = 4+1+12 = 17\] \[(A^2)_{33} = (-1)(-3)+(-2)(-6)+0(0) = 3+12+0 = 15\]

Step 3: Sum the diagonal entries to get the trace. \[\operatorname{trace}(A^2) = 11+17+15 = 43\]

Step 4: Cross-check with the characteristic polynomial. \(\operatorname{trace}(A) = -2+1+0=-1\). The sum of the principal \(2\times2\) minors is \[\begin{vmatrix}-2&2\\2&1\end{vmatrix}+\begin{vmatrix}-2&-3\\-1&0\end{vmatrix}+\begin{vmatrix}1&-6\\-2&0\end{vmatrix} = (-6)+(-3)+(-12) = -21\] and using \(\lambda_1^2+\lambda_2^2+\lambda_3^2 = (\operatorname{trace} A)^2 - 2\sum_{i<j}\lambda_i\lambda_j\), \[\lambda_1^2+\lambda_2^2+\lambda_3^2 = (-1)^2 - 2(-21) = 1+42 = 43\] This matches Step 3.

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