Linear System Solution Analysis

The system of linear equations is given by: $ \begin{aligned} x + y + z &= 6 \\ 2x + 5y + az &= 36 \\ x + 2y + 3z &= b \end{aligned} $ This system can be represented in matrix form $AX = B$, where:

Determinant for Solution Uniqueness

A system of linear equations has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero ($\det(A) \neq 0$).

Let's calculate the determinant of $A$: $ \det(A) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix} $ $ = 1 \cdot (5 \cdot 3 - a \cdot 2) - 1 \cdot (2 \cdot 3 - a \cdot 1) + 1 \cdot (2 \cdot 2 - 5 \cdot 1) $ $ = (15 - 2a) - (6 - a) + (4 - 5) $ $ = 15 - 2a - 6 + a - 1 $ $ = 8 - a $

For a unique solution, $\det(A) \neq 0$. This requires $8 - a \neq 0$, meaning $a \neq 8$.

Case Analysis for $a = 8$

When $a=8$, the determinant is $\det(A) = 8 - 8 = 0$. This indicates that the system does not have a unique solution. It could have infinitely many solutions or no solution.

To determine which case applies, we analyze the augmented matrix $[A|B]$ with $a=8$: $ \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 5 & 8 & 36 \\ 1 & 2 & 3 & b \end{array} \right] $

Perform row reduction:

1. Replace row 2 ($R_2$) with $R_2 - 2R_1$.
2. Replace row 3 ($R_3$) with $R_3 - R_1$.

The matrix becomes:

Swap $R_2$ and $R_3$ for easier calculation:

Replace $R_3$ with $R_3 - 3R_2$:

The last row represents the equation $0 = 24 - 3(b-6)$. Simplifying this gives $0 = 24 - 3b + 18$, which results in $0 = 42 - 3b$.

From $0 = 42 - 3b$, we find $3b = 42$, so $b = 14$.

• If $b=14$, the last row is $0=0$. Since $\text{rank}(A) = 2$ (less than the number of variables), the system has infinitely many solutions.
• If $b \neq 14$, the last row is $0 = \text{non-zero value}$. This is a contradiction, meaning the system has no solution.

Conclusion: For $a=8$, the system never has a unique solution. It has infinitely many solutions only if $b=14$, and no solution otherwise.