[|[arrayccc x+1 & x & x x & x+ & x x & x & x+^2 array]|=(9)/(8)(103 x+81)], then [, ()/(3)…

Question

$$\left|\[\begin{array}{ccc} x+1 & x & x \ x & x+\lambda & x \ x & x & x+\lambda^{2} \end{array}$$\right|=\frac{9}{8}(103 x+81)\], then $$\lambda, \frac{\lambda}{3}$$ are the roots of the equation :

Accepted Answer

$$\left|\[\begin{array}{ccc} x+1 & x & x \ x & x+\lambda & x \ x & x & x+\lambda^{2} \end{array}$$\right|=\frac{9}{8}(103 x+81)\]

Put $x=0$

$$
\begin{aligned}
& \left|\[\begin{array}{ccc} 1 & 0 & 0 \ 0 & \lambda & 0 \ 0 & 0 & \lambda^2 \end{array}\end{aligned}$$\right|=\frac{9}{8} \times 81 \\
\& \lambda\^3=\frac{3\^6}{2\^3}

\]

$$
\begin{aligned}
& \Rightarrow  \lambda=\frac{9}{2} \\
& \Rightarrow \frac{\lambda}{3}=\frac{3}{2}
\end{aligned}
$$

$$
\begin{aligned}
& \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\
& \Rightarrow 4 x^2-24 x+27=0
\end{aligned}
$$

Solutions · 1