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(C) Let the matrix be $A = \begin{bmatrix} 2-x & 2 & 1 \ 1 & 3-x & 1 \ 1 & 2 & 2-x \end{bmatrix}$.Since the matrix $A$ is singular,its determinant m

Vedclass · Accepted Answer

$7$

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(C) Let the matrix be $A = \begin{bmatrix} 2-x & 2 & 1 \ 1 & 3-x & 1 \ 1 & 2 & 2-x \end{bmatrix}$.Since the matrix $A$ is singular,its determinant must be zero,i.e.,$|A| = 0$.$\begin{vmatrix} 2-x & 2 & 1 \ 1 & 3-x & 1 \ 1 & 2 & 2-x \end{vmatrix} = 0$.Applying the column operation $C_1 ightarrow C_1 + C_2 + C_3$,we get:$\begin{vmatrix} 5-x & 2 & 1 \ 5-x & 3-x & 1 \ 5-x & 2 & 2-x \end{vmatrix} = 0$.Taking $(5-x)$ common from the first column:$(5-x) \begin{vmatrix} 1 & 2 & 1 \ 1 & 3-x & 1 \ 1 & 2 & 2-x \end{vmatrix} = 0$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$(5-x) \begin{vmatrix} 1 & 2 & 1 \ 0 & 1-x & 0 \ 0 & 0 & 1-x \end{vmatrix} = 0$.Expanding along the first column:$(5-x) \cdot 1 \cdot [(1-x)(1-x) - 0] = 0$.$(5-x)(1-x)^2 = 0$.Thus,the values of $x$ are $x = 5, 1, 1$.The sum of the values of $x$ is $5 + 1 + 1 = 7$.Therefore,option $C$ is correct.

The sum of the values of $x$ such that the matrix $\begin{bmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix} - x \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is singular,is

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