The sum of the values of $x$ so that the matrix $\left(\begin{array}{lll}2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 &…

The sum of the values of $x$ so that the matrix $\left(\begin{array}{lll}2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2\end{array}\right)-x\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$ is singular, is
  1. 3
  2. 5
  3. 7
  4. 9

Solution

Let a matrix $ =\left[\begin{array}{lll} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{array}\right]-x\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2-x & 2 & 1 \\ 1 & 3-x & 1 \\ 1 & 2 & 2-x \end{array}\right] $ $\because$ Matrix $A$ is a singular. $ \therefore|A|=0 \Rightarrow\left|\begin{array}{ccc} 2-x & 2 & 1 \\ 1 & 3-x & 1 \\ 1 & 2 & 2-x \end{array}\right|=0 $ On applying $C_1 \rightarrow C_1+C_2+C_3$, we get $ \begin{aligned} &\left|\begin{array}{ccc} 5-x & 2 & 1 \\ 5-x & 3-x & 1 \\ 5-x & 2 & 2-x \end{array}\right|=0 \\ & \Rightarrow \quad(5-x)\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 3-x & 1 \\ 1 & 2 & 2-x \end{array}\right|=0 \end{aligned} $ On applying $R_2 \rightarrow R_2-1$ and $R_3 \rightarrow R_3-R_1$, we get $ \begin{aligned} (5-x)\left|\begin{array}{ccc} 1 & 2 & 1 \\ 0 & 1-x & 0 \\ 0 & 0 & 1-x \end{array}\right| & =0 \\ \Rightarrow \quad(5-x)(1-x)^2 & =0 \Rightarrow x=1,1,5 \end{aligned} $ So, sum of the required values of $x$ is 7 . Hence, option (c) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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