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Let A be a square matrix of order 2 or 3 and I will be the identity matrix of the same order. Then the matrix $A − λ I$ is called the characteristic matrix of the matrix A , where $λ$ is a complex number. The determinant of the characteristic matrix is called characteristic determinant of the matrix A which will, of course, be a polynomial of degree 3 in $λ$ . The equation $A − λ I = 0$ is called the characteristic equation of the matrix A and its roots (the values of $λ$ ) are called characteristic roots or eigenvalues. It is also known that every square matrix has its characteristic equation.
The eigenvalues of the matrix $A = 211234 − 1 − 1 − 2$ are,

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2, 1, 1

2, 3, − 2

− 1, 1, 3

None of these

answer is C.

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Detailed Solution

Given that,

A = 211234 − 1 − 1 − 2

.
Since a scalar

λ

is a eigenvalue of the matrix A, if the characteristic equation

A − λ I = 0

is satisfied.
Question Image

Substitute

λ = − 1

in the above equation, whether it is satisfied or not, and we have

λ 3 − 3 λ 2 − λ + 3 = (− 1) 3 − 3 (− 1) 2 − (− 1) + 3 = − 1 − 3 (1) + 1 + 3 = − 1 − 3 + 1 + 3 = 0

It satisfies the equation.
Now, substitute

λ = 1

in the above equation, whether it is satisfied or not, and we have

λ 3 − 3 λ 2 − λ + 3 = (1) 3 − 3 (1) 2 − (1) + 3 = 1 − 3 (1) − 1 + 3 = 1 − 3 − 1 + 3 = 0

It satisfies the equation.
Let the third root of the equation be k. Then,

λ 3 − 3 λ 2 − λ + 3 = (λ − 1) (λ + 1) (λ − k) = λ 2 − 1 (λ − k) = λ 3 − λ 2 k − λ + k

Equating on both sides, we get

k = 3

.
Therefore, the eigenvalues are

− 1, 1, 3

.
Hence, the correct option is 3.

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