To find the value of k for which the given system of equations has a solution, let's analyze the system using matrix methods.

The given system of equations can be written in matrix form as:

| 1 1 1 | | x | | 1 |
| 2 1 4 | * | y | = | k |
| 4 1 10| | z | | 2 |

We can represent the coefficient matrix as A, the variable matrix as X, and the constant matrix as B. Therefore, the system can be written as AX = B.

We can then find the inverse of matrix A, denoted as A^(-1), which exists if and only if the determinant of matrix A is non-zero.

To find the determinant of matrix A, we can use the rule of Sarrus or any other suitable method. However, in this case, the determinant can be calculated easily using the Laplace expansion along the first row.

Determinant of matrix A:
| 1 1 1 |
| 2 1 4 |
| 4 1 10|

= 1 * (1 * 10 - 4 * 1) - 1 * (2 * 10 - 4 * 4) + 1 * (2 * 1 - 1 * 4)
= 1 * 6 - 1 * (-12) + 1 * (-2)
= 6 + 12 - 2
= 16

Since the determinant of matrix A is non-zero (16 ≠ 0), we can conclude that the inverse of matrix A, A^(-1), exists.

Now, we can find the inverse of matrix A using the formula:

A^(-1) = (adj(A)) / (det(A))

where adj(A) represents the adjugate of matrix A.

The adjugate of matrix A can be found by taking the transpose of the matrix of cofactors of A.

Cofactor matrix of A:
| 1 -2 1 |
| -2 4 -2|
| 1 -2 1 |

Transpose of the cofactor matrix:
| 1 -2 1 |
| -2 4 -2|
| 1 -2 1 |

Now, we can find the inverse of matrix A:

A^(-1) = (1/16) * | 1 -2 1 |
| -2 4 -2|
| 1 -2 1 |

Multiplying both sides of the equation AX = B by A^(-1), we get:

X = A^(-1) * B

Substituting the values of A^(-1) and B, we have:

| x | | 1 -2 1 | | 1 |
| y | = (1/16) * | -2 4 -2| * | k |
| z | | 1 -2 1 | | 2 |

Simplifying the matrix multiplication:

| x | | 1/16 * (1 - 2 + 1) | | 1 |
| y | = | 1/16