Maths

Question 3

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  • 22 September, 2026 at 1:07 am #11715
    shahkb4
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    Let $\alpha, \beta \in R$ be such that the system of linear equations $x+2 y+z=5,2 x+y+\alpha z=5,8 x+4 y+\beta z=18$ has no

    22 September, 2026 at 1:07 am #11716
    shahkb4
    Keymaster

    Solution

    for no solution $\Delta=\left|\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta\end{array}\right|=4 \alpha-\beta=0 \Rightarrow \beta=4 \alpha$
    Second eq. $2 \mathrm{x}+\mathrm{y}+\alpha \mathrm{z}=5$
    third eq. $8 x+4 y+4 \alpha z=18 \Rightarrow 2 x+y+\alpha z=\frac{9}{2}$ both are parallel planes $\therefore$ no solution
    [CHAPTER M26 MATRICES TOPIC T2 ]

    22 September, 2026 at 1:07 am #11717
    shahkb4
    Keymaster

    Answer

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