Problem Statement

Consider a 3×3 matrix A with parameter k.
The eigenvalues of the matrix are given as −2, 1, and 2.
Find the value of k.

Concept Used

The eigenvalues of a matrix are the roots of its characteristic equation
det(A − λI) = 0.

By Vieta’s formulas:

• Sum of eigenvalues = Trace of A
• Product of eigenvalues = Determinant of A

Step 1: Use the Trace Condition

The trace of matrix A is the sum of its diagonal elements.

Trace(A) = 1 + k + (−1) = k

Given eigenvalues: −2, 1, and 2

Sum of eigenvalues = −2 + 1 + 2 = 1

Equating trace with sum of eigenvalues:

k = 1

Step 2: Verify Using Determinant

Substitute k = 1 into matrix A:

A =
| 1  0  1 |
| 0  1  0 |
| 3  0 -1 |

Compute the determinant:

det(A) = 1·det
| 1 0 |
| 0 −1 |
− 0 + 1·det
| 0 1 |
| 3 0 |

det(A) = 1(−1) + 1(−3) = −4

Product of given eigenvalues:

(−2)(1)(2) = −4

✔ Determinant condition is satisfied.

Step 3: Characteristic Polynomial Check

For k = 1, the characteristic polynomial expands to:

(λ + 2)(λ − 1)(λ − 2) = 0

Hence, eigenvalues are exactly −2, 1, and 2.

Option Analysis

k Value	Trace Matches?	Determinant Matches?
0	No	No
1	Yes	Yes
2	No	No
3	No	No

Final Answer

✔ The correct value of k is 1.