Matrices and Determinants TG EAPCET Previous Year Chapters wise Questions

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To solve the given determinant equation, we start by simplifying the expression:$\begin{vmatrix} x-2 & 3(x-1) & 5(x-1) \\ x-4 & 3(x-3) & 5(x-5) \\ x-8 & 3(x-9) & 5(x-25) \end{vmatrix}=0$This simplifies to:$\begin{vmatrix} x-2 & x-1 & x-1 \\ x-4 & x-3 & x-5 \\ x-8 & x-9 & x-25 \end{vmatrix}=0$Next, perform column operations to simplify the determinant. Subtract the second column from the first:$C_1 \rightarrow C_1 - C_2 \Rightarrow \begin{vmatrix} -1 & x-1 & x-1 \\ -1 & x-3 & x-5 \\ 1 & x-9 & x-25 \end{vmatrix}=0$Now, apply row operations:Add the third row to the first row: $R_1 \rightarrow R_1 + R_3$Add the third row to the second row: $R_2 \rightarrow R_2 + R_3$This results in:$\begin{vmatrix} 0 & 2x-10 & 2x-26 \\ 0 & 2x-12 & 2x-30 \\ 1 & x-9 & x-25 \end{vmatrix}=0$To solve, expand the determinant by the first column:$(2x-30)(2x-10)-(2x-12)(2x-26)=0$This simplifies to:$2(x-15) \cdot 2(x-5)-2(x-6) \cdot 2(x-13)=0$Further simplifying, we have:$(x-15)(x-5)-(x-6)(x-13)=0$Solving this, we calculate:$x^2-20x+75-(x^2-19x+78)=0$This leads to:$-x-3=0 \Rightarrow x=-3$Therefore, $x=-3$ satisfies the equation $x^2+2x-3=0$.