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

Question

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

UpStudy AI · Accepted Answer

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.
For the system to have no solution, $$ a $$ must be 1 and $$ b $$ must not be -4.

What should be the values of a and b so that the system
solution.

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What should be the values of a and b so that the system solution.