If \(\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then…
If \(\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) has no inverse, then the real value of \(x\) is
2
3
0
1
Solution
It is given that matrix \(A=\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]\) have no inverse, so
\(\begin{array}{rlrl}
& |A|=0 \Rightarrow\left|\begin{array}{ccc}
1 & -1 & x \\
1 & x & 1 \\
x & -1 & 1
\end{array}\right| & =0 \\
\Rightarrow \quad 1(x+1)+1(1-x)+x\left(-1-x^2\right) & =0 \\
\Rightarrow \quad x+1+1-x-x-x^3 & =0 \\
\Rightarrow x^3+x-2=0 \Rightarrow(x-1)\left(x^2+x+2\right) & =0
\end{array}\)
Either \(x=1\) or \(x^2+x+2=0\)
But discriminant of quadratic equation \(x^2+x+2=0\) is negative so no real roots.
\(\therefore|A|=0 \Rightarrow x=1\)
Hence, option (d) is correct