The eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.x1&nbsp;′=2x1+x2−x3x2&nbsp;′=−4x1−3x2−x3x3&nbsp;′=4x1+4x2+2x3How to find the eigenvalues by using the fact that&nbsp;|A−λI|=0, how do it by inspection?Imagine you have the matrix,&nbsp;A=(2−1−1−12−1−1−12)By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

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The eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.x1&amp;nbsp;′=2x1+x2−x3x2&amp;nbsp;′=−4x1−3x2−x3x3&amp;nbsp;′=4x1+4x2+2x3How to find the eigenvalues by using the fact that&amp;nbsp;|A−λI|=0, how do it by inspection?Imagine you have the matrix,&amp;nbsp;A=(2−1−1−12−1−1−12)By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

Elois Puryear · Accepted Answer

Let A - the matrix of coefficients, then&amp;nbsp;(A⋅x)k=λxk.&amp;nbsp;Add the third and the second row.&amp;nbsp;So&amp;nbsp;x2+x3=λ(x2+x3), thus&amp;nbsp;λ=1&amp;nbsp;or&amp;nbsp;x2+x3=0. Since the trace of A equals 1, the sum of other two eigenvalues is 0.&amp;nbsp;For other eigenvalues, let&amp;nbsp;x3=−x2, giving&amp;nbsp;2x1+2x2=λx1&amp;nbsp;and&amp;nbsp;−4x1−2x2=λx2. The characteristic polynomial is easily found as&amp;nbsp;λ2+4=0.

Terry Ray · Accepted Answer

The first eigenvalue may be inferred from the fact that A-I has two equal rows and two equal columns.The additional two eigenvalues are then provided to you by the trace and determinant.

alenahelenash · Accepted Answer

The determinant of a matrix is the sum of the eigenvalues., so in this case&amp;nbsp;λ1λ2λ3=4. The sum of the diagonals is the sum of the eigenvalues so&amp;nbsp;λ1+λ2+λ3=2−3+2=1.&amp;nbsp;[−110]T&amp;nbsp;is an eigenvector with eigenvalue 1, then&amp;nbsp;λ1=1&amp;nbsp;so&amp;nbsp;λ2λ3=4&amp;nbsp;and&amp;nbsp;λ2+λ3=0. Thus&amp;nbsp;λ2,λ3=±2−1.In general there is no easy way to spot eigenvalues of&amp;nbsp;3×3&amp;nbsp;matrices.With&amp;nbsp;2×2&amp;nbsp;matrices you can use the determinant and trace. The fact that these are the only non-trivial coefficients of the characteristic polynomial is the primary cause of why this works (and thus determine the eigenvalues). Other characteristic polynomial coefficients aren&#039;t as simple to use or recall when working with larger matrices, which necessitates the employment of more invariants (i.e., coefficients of the characteristic polynomial).

Finding eigenvalues by inspection? I need to solve the following problem, In this problem, t

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