First slide

System of linear equations

Question

$For which of the following ordered pairs (μ, δ), the system of linear equations$
$x + 2 y + 3 z = 1, 3 x + 4 y + 5 z = μ a n d 4 x + 4 y + 4 z = δ$ $i s i n c o n s i s t e n t ?$

Very difficult

A

$(4, 6)$
B

$(1, 0)$
C

$(3, 4)$
D

$(4, 3)$

Solution

$Note D = |\begin{array}{l} 3 & 4 & 5 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{array}| (R_{3} \to R_{3} - 2 R_{1} + 3 R_{2}) = |\begin{array}{l} 3 & 4 & 5 \\ 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}| = 0$
$\begin{array}{l} Now let P_{3} = 4 x + 4 y + 4 z - δ = 0 . If the system has solutions it will have infinite \\ solution, so P_{3} = α P_{1} + β P_{2} \end{array}$
$\begin{array}{l} Hence 3 α + β = 4 & 4 α + 2 β = 4 \Rightarrow α = 2 & β = - 2 \\ So for infinite solution 2 μ - 2 = δ \Rightarrow for 2 μ \neq δ + 2 \end{array}$
system inconsistent

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