First slide

System of linear equations

Question

The system of equations $(\begin{array}{rrr} 3 - 2 1 \\ 5 - 8 9 \\ 2 1 a \end{array}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{r} b \\ 3 \\ - 1 \end{array})$ has no solution if a and b are

Moderate

A

$a = - 3, b \neq 1 / 3$
B

$a = 2 / 3, b \neq 1 / 3$
C

$a \neq 1 / 4, b = 1 / 3$
D

$a \neq - 3, b \neq 1 / 3$

Solution

The augmented matrix is given by
$(A ∣ B) = (\begin{array}{rrrr} 3 - 2 1 b \\ 5 - 8 9 3 \\ 2 1 a - 1 \end{array})$
Applying $R_{1} \to 2 R_{1} - R_{2}$ , we get
$(A ∣ B) ~ (\begin{array}{rrrc} 1 & 4 & - 7 & 2 b - 3 \\ 5 & - 8 & 9 & 3 \\ 2 & 1 & a & - 1 \end{array})$
Applying $R_{2} \to R_{2} - 5 R_{1}, R_{3} \to R_{3} - 2 R_{1}$ , we get
$(A ∣ B) ~ (\begin{array}{cccc} 1 & 4 & - 7 & 2 b - 3 \\ 0 & - 28 & 44 & 18 - 10 b \\ 0 & - 7 & a + 14 & 5 - 4 b \end{array})$
The system of equations will have no solution If
$\frac{- 28}{- 7} = \frac{44}{a + 14} \neq \frac{18 - 10 b}{5 - 4 b}$
$\Rightarrow a + 14 = 11$ and $20 - 16 b \neq 18 - 10 b$
$\Rightarrow a = - 3$ and $b \neq - 1 / 3$

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