8. What should be the values of \( a \) and \( b \) so that the system \( \left\{\begin{array}{l}x+y+z=0 \\ 3 x-a y-z=4 \\ x+5 y+5 z=b\end{array}\right. \) has no solution.

To determine the values of \( a \) and \( b \) for which the system \[ \begin{cases} x + y + z = 0 \\ 3x - a y - z = 4 \\ x + 5y + 5z = b \end{cases} \] has no solution, we can use the concept of **inconsistent systems** in linear algebra. An inconsistent system occurs when there is no set of values for \( x \), \( y \), and \( z \) that satisfies all three equations simultaneously. ### Step 1: Form the Augmented Matrix The system can be represented by the augmented matrix: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 3 & -a & -1 & | & 4 \\ 1 & 5 & 5 & | & b \end{bmatrix} \] ### Step 2: Calculate the Determinant For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. However, if the determinant is zero, the system could either have infinitely many solutions or no solution. We need to find when the system is inconsistent (i.e., no solution). The coefficient matrix is: \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & -a & -1 \\ 1 & 5 & 5 \end{bmatrix} \] The determinant of \( A \) is: \[ \text{det}(A) = 1(-a \cdot 5 - (-1) \cdot 5) - 1(3 \cdot 5 - (-1) \cdot 1) + 1(3 \cdot 5 - (-a) \cdot 1) \] \[ = (-5a + 5) - 16 + (15 + a) = -4a + 4 \] Setting the determinant to zero for a potential inconsistent system: \[ -4a + 4 = 0 \implies a = 1 \] ### Step 3: Check for Inconsistency Substituting \( a = 1 \) back into the augmented matrix: \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 3 & -1 & -1 & | & 4 \\ 1 & 5 & 5 & | & b \end{bmatrix} \] Performing row operations: 1. \( R2 \leftarrow R2 - 3R1 \): \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -4 & -4 & | & 4 \\ 1 & 5 & 5 & | & b \end{bmatrix} \] 2. \( R3 \leftarrow R3 - R1 \): \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -4 & -4 & | & 4 \\ 0 & 4 & 4 & | & b \end{bmatrix} \] 3. \( R3 \leftarrow R3 + R2 \): \[ \begin{bmatrix} 1 & 1 & 1 & | & 0 \\ 0 & -4 & -4 & | & 4 \\ 0 & 0 & 0 & | & b + 4 \end{bmatrix} \] The last row \( 0x + 0y + 0z = b + 4 \) must satisfy \( b + 4 = 0 \) for consistency. If \( b \neq -4 \), the system is **inconsistent**. ### Conclusion The system has **no solution** when: \[ a = 1 \quad \text{and} \quad b \neq -4 \] **Answer:** All real numbers a = 1 together with b different from –4. In other words, a = 1 and b ≠ −4