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

Question

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

UpStudy AI · Accepted Answer

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
The system has no solution when $$ a = 1 $$ and $$ b \neq -4 $$.

What should be the values of and so that the system has no
solution.

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What should be the values of and so that the system has no solution.