System of linear equations

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Question

If the system of linear equations $x + y + z = 6, x + 2 y + 3 z = 10, 3 x + 2 y + λ z = μ$ has more than two solutions, then $λ^{3} + μ^{3} =$

Moderate

Solution

Augmented matrix $= |\begin{matrix} 1 \\ 1 \\ 3 \end{matrix} \begin{matrix} 1 & 1 & 6 \\ 2 & 3 & 10 \\ 2 & λ & μ \end{matrix}| a p p l y R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - 3 R_{1}$
$\begin{array}{l} ~ |\begin{matrix} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & - 1 & λ - 3 & μ - 18 \end{matrix}| a p p l y R_{3} \to R_{3} + R_{2} \\ ~ |\begin{matrix} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & λ - 1 & μ - 14 \end{matrix}| \end{array}$
The system has more than 2 solutions.
$\begin{array}{l} \Rightarrow λ - 1 = 0, μ - 14 = 0 \\ \Rightarrow λ = 1, μ = 14 \\ ∴ λ^{2} + μ^{3} = 1^{3} + 14^{3} = 2745 \end{array}$

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System of linear equations

Remember concepts with our Masterclasses.

If the system of linear equations x+y+z=6,x+2y+3z=10,3x+2y+λz=μ has more than two solutions, then λ3+μ3=

Augmented matrix =113 11623102λμ apply R2→R2−R1, R3→R3−3R1 ~111601240-1λ-3μ-18 apply R3→R3+R2~1116012400λ-1μ-14The system has more than 2 solutions.⇒λ-1=0, μ-14=0⇒λ=1, μ=14∴λ2+μ3=13+143=2745

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If the system of linear equations $x + y + z = 6, x + 2 y + 3 z = 10, 3 x + 2 y + λ z = μ$ has more than two solutions, then $λ^{3} + μ^{3} =$