Given the following matrix where c and r are arbitrary real numbers and 5.5

Question

Given the following matrix

$A = \begin{bmatrix} 1 & 2 & r \\ c & 1 & 7 \\ c & 1 & 7 \end{bmatrix},$

where c and r are arbitrary real numbers and 5.5 < r ≤ 6.5, and the fact that $\lambda_1$ = 3 is one of the eigenvalues, is it possible to determine the other two eigenvalues? If so, compute them and give reasons for your answer.

Solution

Characteristic Equation of A is

$\begin{align*} \left det | A - \lambda I | = f(\lambda) = 0 \end{align*}$

A is a $3 \times 3$ matrix. Thus, it can have at most three eigenvalues.

Since rows R2 and R3 are the same, we know that the determinant of the matrix is 0. We know that the product of eigenvalues equals the determinant of the matrix. Thus, at least one of the eigenvalues should be zero.

We also know that the sum of the diagonal elements of the matrix is equal to the sum of its eigenvalues. So far, we know that there are three eigenvalues and that one of them is zero and the other is 3.

Thus,

$\begin{align*} \left \lambda_1 + \lambda_2 + \lambda_3 = 1 + 1 + 7 \end{align*}$

$\begin{align*} \left \Rightarrow 3 + 0 + \lambda_3 = 1 + 1 + 7 \end{align*}$

$\begin{align*} \left \Rightarrow \lambda_3 = 6 \end{align*}$

Thus, the A has the eigenvalues are 0, 3 and 6.

Question

Solution

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