Let `A=[{:(1,2,1),(0,1,-1),(3,1,1):}]`. Find the sum of all the value of `lamda` for which there exists a column vertor `Xne0` such that `AX=lamdaX`.

Explanation

To find the sum of all values of \( \lambda \) for which there exists a column vector \( X \neq 0 \) such that \( AX = \lambda X \), we can follow these steps:

### Step 1: Understand the Equation
We start with the equation \( AX = \lambda X \). Rearranging this gives us:
\[ AX - \lambda X = 0 \]
This can be rewritten as:
\[ (A - \lambda I)X = 0 \]
where \( I \) is the identity matrix.

### Step 2: Characteristic Equation
For the equation \( (A - \lambda I)X = 0 \) to have non-trivial solutions (i.e., \( X \neq 0 \)), the determinant of the matrix \( (A - \lambda I) \) must be zero:
\[ \text{det}(A - \lambda I) = 0 \]
This leads us to the characteristic polynomial of the matrix \( A \).

### Step 3: Calculate the Trace of Matrix A
The sum of all eigenvalues (which are the values of \( \lambda \)) is equal to the trace of the matrix \( A \). The trace is defined as the sum of the diagonal elements of the matrix.

Given the matrix:
\[ A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & 1 & 1 \end{pmatrix} \]
we can calculate the trace:
\[ \text{Trace}(A) = 1 + 1 + 1 = 3 \]

### Step 4: Conclusion
Thus, the sum of all values of \( \lambda \) for which there exists a non-zero vector \( X \) such that \( AX = \lambda X \) is:
\[ \text{Sum of all } \lambda = \text{Trace}(A) = 3 \]

### Final Answer
The sum of all values of \( \lambda \) is \( \boxed{3} \).

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