The eigenvalues of $A = \left( \begin{array} { c c c }
1 & - 12 & - 14 \
1 & 2 & - 3 \
1 & 1 & - 2
\end{array} \right)$ is
A) $1 , \pm 5 i$
B) $- 1 , \pm 5 i$
C) $1,4 , - 6$
D) $1 , - 4,6$
E) none of the above

Question

The eigenvalues of $A = \left( \begin{array} { c c c } 
1 & - 12 & - 14 \
1 & 2 & - 3 \
1 & 1 & - 2
\end{array} \right)$ is
A) $1 , \pm 5 i$
B) $- 1 , \pm 5 i$
C) $1,4 , - 6$
D) $1 , - 4,6$
E) none of the above

Quizplus · Accepted Answer

To find the eigenvalues of matrix $A$, we need to solve the characteristic equation $\det(A - \lambda I) = 0$, where $I$ is the identity matrix and $\lambda$ are the eigenvalues. The characteristic polynomial of $A$ is obtained by calculating the determinant of $A - \lambda I$, which results in a cubic polynomial. Solving this polynomial will give us the eigenvalues of $A$. For this specific matrix, solving the characteristic equation yields the eigenvalues $1, \pm 5i$, indicating complex eigenvalues with real part 1 and imaginary parts $\pm 5$.

The Eigenvalues Of A=(1−12−1412−311−2)A = \left( \begin{array} { c c c } 1 & - 12 & - 14 \\1 & 2 & - 3 \\1 & 1 & - 2\end{array} \right)A=​111​−1221​−14−3−2​​

The Eigenvalues Of $A = \left( \begin{array} { c c c } 1 & - 12 & - 14 \\1 & 2 & - 3 \\1 & 1 & - 2\end{array} \right)$