91Ó°ÊÓ

To determine the eigenvalues of a matrix, we must solve the characteristic equation. For a given matrix \(A\), the characteristic equation is expressed as \(\det(A - \lambda I) = 0\). Here, \(\lambda\) symbolizes eigenvalues we are solving for, and \(I\) is the identity matrix.

In this exercise, for matrix \(A = \begin{array}{rrr}3 & 5 & 2 \ -8 & -11 & -4 \ 10 & 11 & 3\end{array}\), we calculate the determinant of \(A - \lambda I\). This gives us a polynomial equation in terms of \(\lambda\).

Upon subtracting \(\lambda I\) from \(A\), we get:

\(A - \lambda I = \begin{array}{rrr} 3 - \lambda & 5 & 2 \ -8 & -11 - \lambda & -4 \ 10 & 11 & 3 - \lambda \end{array}\).

Now, we compute the determinant of this square matrix which results in a polynomial. Solving the determinant results in the characteristic polynomial which equates to zero when expanded:
\[-(\lambda + 3)(\lambda + 2)(\lambda + 5) = 0\]\
The solutions to this polynomial (roots) give us the eigenvalues. We thus have, \(\lambda_1 = -3\) , \(\lambda_2 = -2\) , and \(\lambda_3 = -5\).

The determinant of a matrix is a special number that summarizes certain properties of the matrix. For a general square matrix \(A\), the determinant is often denoted as \(\det(A)\) or \(|A|\).

In the context of finding eigenvalues, we need to compute the determinant of the matrix \(A - \lambda I\). The resulting determinant is a polynomial in terms of \(\lambda\).

Here's how we computed it in this exercise: Starting with the matrix:

\(A - \lambda I = \begin{array}{rrr} 3 - \lambda & 5 & 2 \ -8 & -11 - \lambda & -4 \ 10 & 11 & 3 - \lambda \end{array}\).\

We proceed to calculate the determinant by expanding the matrix:
\[\det(A - \lambda I) = (3 - \lambda) \det \begin{array}{rrr} -11 - \lambda & -4 \ 11 & 3 - \lambda \end{array} - 5 \det \begin{array}{rrr} -8 & -4 \ 10 & 3 - \lambda \end{array} + 2 \det \begin{array}{rrr} -8 & -11 - \lambda \ 10 & 11 \end{array}\].\

By further expanding and simplifying the determinants of the sub-matrices, we find our characteristic polynomial, which factors to:
\[(\lambda + 3)(\lambda + 2)(\lambda + 5).\]\
Setting this polynomial to zero allows us to solve for the eigenvalues.

Linear equation systems play a significant role in finding the eigenvectors corresponding to the eigenvalues. An eigenvector corresponding to an eigenvalue \(\lambda\) satisfies the linear equation system \((A - \lambda I)\bm{v} = \bm{0}\), where \(\bm{v}\) is the eigenvector and \(\bm{0}\) is the zero vector.

From our exercise, we locate the eigenvectors for each eigenvalue as follows:

For \(\lambda = -3\):
Solving \( (A + 3I)\bm{v} = \bm{0} \) yields the system:
\[(\begin{array}{rrr} 6 & 5 & 2 \ -8 & -8 & -4 \ 10 & 11 & 6 \end{array})\bm{v} = \bm{0}\].\ \
Solving which gives us the eigenvector \(\bm{v_1} = \begin{array}{r} -1/3 \ 1 \ 0 \end{array}\).

For \(\lambda = -2\):
Solving \( (A + 2I)\bm{v} = \bm{0} \) yields the system:
\[(\begin{array}{rrr} 5 & 5 & 2 \ -8 & -9 & -4 \ 10 & 11 & 5 \end{array})\bm{v} = \bm{0}\].\
Solving which gives us the eigenvector \(\bm{v_2} = \begin{array}{r} -1 \ 1 \ -1 \end{array}\).

For \(\lambda = -5\):
Solving \( (A + 5I)\bm{v} = \bm{0} \) yields the system:
\[(\begin{array}{rrr} 8 & 5 & 2 \ -8 & -6 & -4 \ 10 & 11 & 8 \end{array}) \bm{v} = \bm{0}\].\ \
Solving which gives us the eigenvector \(\bm{v_3} = \begin{array}{r} 0.5 \ -1 \ 1 \end{array}\).