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

To determine the values of \( a \) and \( b \) for which the system of equations \[ \begin{cases} x + y + z = 0 \\ 3x - a y - z = 4 \\ x + 5y + 5z = b \end{cases} \] has **no solution**, we need to analyze the system for consistency. ### Step 1: Form the Augmented Matrix The augmented matrix for the system is: \[ \begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 3 & -a & -1 & | & 4 \\ 1 & 5 & 5 & | & b \\ \end{pmatrix} \] ### Step 2: Perform Row Operations 1. **Eliminate \( x \) from the second and third equations:** - **Row2'** = Row2 - 3·Row1: \[ 3x - a y - z - 3(x + y + z) = 4 - 0 \implies - (a + 3)y - 4z = 4 \] So, Row2' becomes: \[ - (a + 3)y - 4z = 4 \] - **Row3'** = Row3 - Row1: \[ x + 5y + 5z - (x + y + z) = b - 0 \implies 4y + 4z = b \] So, Row3' becomes: \[ 4y + 4z = b \] 2. **Rewrite the simplified system:** \[ \begin{cases} - (a + 3)y - 4z = 4 \\ 4y + 4z = b \\ \end{cases} \] ### Step 3: Analyze for Inconsistency For the system to have **no solution**, these two equations must be **parallel but have different constants**, meaning their coefficients are proportional, but the constants are not. 1. **Check proportionality of coefficients:** \[ \frac{-(a + 3)}{4} = \frac{-4}{4} \implies -(a + 3) = -4 \implies a + 3 = 4 \implies a = 1 \] 2. **Ensure the constants are not in the same ratio:** When \( a = 1 \), the first equation becomes: \[ -4y - 4z = 4 \implies y + z = -1 \] The second equation becomes: \[ 4y + 4z = b \implies y + z = \frac{b}{4} \] For these to be inconsistent: \[ -1 \neq \frac{b}{4} \implies b \neq -4 \] ### Conclusion The system has **no solution** if and only if: - \( a = 1 \) - \( b \neq -4 \) ### Final Answer All real numbers a = 1 and b different from –4. In other words, a = 1 and b ≠ −4.