$$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$, then $\lambda, \frac{\lambda}{3}$ are the roots of the equation :
Solution
$$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$
<br/><br/>Put $x=0$
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & \lambda & 0 \\
0 & 0 & \lambda^2
\end{array}\right|=\frac{9}{8} \times 81 \\\\
& \lambda^3=\frac{3^6}{2^3}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \lambda=\frac{9}{2} \\\\
& \Rightarrow \frac{\lambda}{3}=\frac{3}{2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\
& \Rightarrow 4 x^2-24 x+27=0
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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