KCET · Maths · Matrices
If \(\left[\begin{array}{ccr}1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1\end{array}\right]\) is singular, then the value of \(x\) is
- A 2
- B 3
- C 1
- D 0
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
Let \(A=\left[\begin{array}{ccc}1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1\end{array}\right]\) if the matrix \(A\) is singular, then \(|A|=0\)
\[
\left|\begin{array}{ccc}
1 & 2 & -1 \\
1 & x-2 & 1 \\
x & 1 & 1
\end{array}\right|=0
\]
Expand with respect to \(R_{1}\)
\(\Rightarrow 1(x-2-1)-2(1-x)-1\left(1-x^{2}+2 x\right)=0\)
\(\Rightarrow \quad x-3-2+2 x-1+x^{2}-2 x=0\)
\(\Rightarrow \quad x^{2}+x-6=0\)
\(\Rightarrow \quad(x-2)(x+3)=0\)
\(\Rightarrow \quad x=2,-3\)
\[
\left|\begin{array}{ccc}
1 & 2 & -1 \\
1 & x-2 & 1 \\
x & 1 & 1
\end{array}\right|=0
\]
Expand with respect to \(R_{1}\)
\(\Rightarrow 1(x-2-1)-2(1-x)-1\left(1-x^{2}+2 x\right)=0\)
\(\Rightarrow \quad x-3-2+2 x-1+x^{2}-2 x=0\)
\(\Rightarrow \quad x^{2}+x-6=0\)
\(\Rightarrow \quad(x-2)(x+3)=0\)
\(\Rightarrow \quad x=2,-3\)
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