System of linear equations

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If the system of linear equations $4 x + k y + 2 z = 0, k x + 4 y + z = 0, 2 x + 2 y + z = 0$ possess a non-zero solution, so that k has two values $k_{1}, k_{2}$ then $\frac{k_{1}^{3} + k_{2}^{3}}{3^{3}} =$

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Solution

System has non-zero solution $\Rightarrow Δ = 0$
$\Rightarrow |\begin{matrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{matrix}| = 0 \Rightarrow 4 (4 - 2) - k (k - 2) + 2 (2 k - 8) = 0 \Rightarrow k^{2} - 6 k + 8 = 0 \Rightarrow k = 2, 4 L e t k_{1} = 2 a n d k_{2} = 4 . T h e n \frac{{k_{1}}^{3} + {k_{2}}^{3}}{3^{3}} = \frac{8 + 64}{27} = \frac{72}{27} = 2.6667$

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System of linear equations

Remember concepts with our Masterclasses.

If the system of linear equations 4x+ky+2z=0,kx+4y+z=0,2x+2y+z=0 possess a non-zero solution, so that k has two values k1,k2then k13+k2333=

System has non-zero solution⇒Δ=0 ⇒4k2k41221=0 ⇒44-2-kk-2+22k-8=0 ⇒k2−6k+8=0 ⇒k=2,4 Let k1=2 and k2=4. Then k13+k2333=8+6427=7227=2.6667

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If the system of linear equations $4 x + k y + 2 z = 0, k x + 4 y + z = 0, 2 x + 2 y + z = 0$ possess a non-zero solution, so that k has two values $k_{1}, k_{2}$ then $\frac{k_{1}^{3} + k_{2}^{3}}{3^{3}} =$