Step-by-step Solution:

The given system of linear equations is:

1. $x - 2y + 3z = -1$
2. $-x + y - 2z = k$
3. $x - 3y + 4z = 1$

This system can be written in matrix form as $AX = B$, where: $A = \begin{pmatrix} 1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} -1 \\ k \\ 1 \end{pmatrix}$

Analysis of Statement - 1

Statement - 1 : The system of equations has no solution for $k \ne 3$.

To determine the nature of the solution, we first calculate the determinant of the coefficient matrix, $\det(A)$.

$\det(A) = \left| {\begin{matrix} 1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4 \\ \end{matrix} } \right|$ $= 1(1 \cdot 4 - (-2)(-3)) - (-2)(-1 \cdot 4 - (-2) \cdot 1) + 3(-1 \cdot (-3) - 1 \cdot 1)$ $= 1(4 - 6) + 2(-4 + 2) + 3(3 - 1)$ $= 1(-2) + 2(-2) + 3(2)$ $= -2 - 4 + 6 = 0$

Since $\det(A) = 0$, the system will have either infinitely many solutions or no solution.

To find the condition for consistency, we can use Gaussian elimination on the augmented matrix $[A|B]$.

$[A|B] = \left[ {\begin{matrix} 1 & -2 & 3 & | & -1 \\ -1 & 1 & -2 & | & k \\ 1 & -3 & 4 & | & 1 \\ \end{matrix} } \right]$

Apply row operations: $R_2 \to R_2 + R_1$ $R_3 \to R_3 - R_1$

$\sim \left[ {\begin{matrix} 1 & -2 & 3 & | & -1 \\ 0 & -1 & 1 & | & k-1 \\ 0 & -1 & 1 & | & 2 \\ \end{matrix} } \right]$

Next, apply $R_3 \to R_3 - R_2$:

$\sim \left[ {\begin{matrix} 1 & -2 & 3 & | & -1 \\ 0 & -1 & 1 & | & k-1 \\ 0 & 0 & 0 & | & 2 - (k-1) \\ \end{matrix} } \right] = \left[ {\begin{matrix} 1 & -2 & 3 & | & -1 \\ 0 & -1 & 1 & | & k-1 \\ 0 & 0 & 0 & | & 3-k \\ \end{matrix} } \right]$

The last row represents the equation $0 \cdot x + 0 \cdot y + 0 \cdot z = 3 - k$, which simplifies to $0 = 3 - k$.

• If $3 - k \ne 0$ (i.e., $k \ne 3$), the equation becomes $0 = (\text{non-zero value})$, which is a contradiction. In this case, the system is inconsistent and has no solution.
• If $3 - k = 0$ (i.e., $k = 3$), the equation becomes $0 = 0$, which is always true. The system has infinitely many solutions.

Therefore, the system of equations has no solution for $k \ne 3$.

Conclusion: Statement - 1 is True.

Analysis of Statement - 2

Statement - 2 : The determinant $\left| {\begin{matrix} 1 & 3 & { - 1} \\ { - 1} & { - 2} & k \\ 1 & 4 & 1 \\ \end{matrix} } \right| \ne 0$, for $k \ne 3$.

Let's calculate the value of the determinant:

$D = \left| {\begin{matrix} 1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1 \\ \end{matrix} } \right|$ $= 1(-2 \cdot 1 - k \cdot 4) - 3(-1 \cdot 1 - k \cdot 1) + (-1)(-1 \cdot 4 - (-2) \cdot 1)$ $= 1(-2 - 4k) - 3(-1 - k) - 1(-4 + 2)$ $= -2 - 4k + 3 + 3k - 1(-2)$ $= -2 - 4k + 3 + 3k + 2$ $= (-2 + 3 + 2) + (-4k + 3k)$ $= 3 - k$

The statement says that this determinant is not equal to zero for $k \ne 3$. If $D = 0$, then $3 - k = 0$, which means $k = 3$. So, the determinant is zero only when $k=3$. For any value $k \ne 3$, the determinant $D = 3-k$ will be non-zero.

Conclusion: Statement - 2 is True.

Relationship between Statement - 1 and Statement - 2

• We found that Statement 1 is true. The condition for the system to have no solution is $k \ne 3$. The mathematical reason is that $\det(A) = 0$ and the system becomes inconsistent if $3-k \ne 0$.

• We found that Statement 2 is also true. It states that a certain determinant is non-zero for $k \ne 3$. This determinant evaluates to $3-k$. So, Statement 2 is equivalent to the condition $3-k \ne 0$ for $k \ne 3$.

• The condition $3-k \ne 0$ is precisely the reason why the system in Statement 1 is inconsistent (has no solution), given that we already established $\det(A)=0$. Statement 2 provides this crucial condition.

Therefore, Statement 2 is a correct explanation for Statement 1.

Final Choice: Statement - 1 is True, Statement - 2 is True; Statement - 2 is a correct explanation for Statement - 1. This corresponds to option A.