Practicing Success
The value(s) of x, for which the matrix $A=\left[\begin{array}{ccc}x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right]$ is singular, is/are : |
1 -1, 2 ±1 1, 2 |
-1, 2 |
$A=\left[\begin{array}{ccc}x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right]$ we need to find x for |A| = 0 $|A|=\left|\begin{array}{ccc}x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right|$ ⇒ Using operation R1 → R1 + R2 + R3 $|A|=\left|\begin{array}{ccc}x+1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right|$ ⇒ So $|A|=(x+1)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right|$ ⇒ Using operation R2 → R2 - R1 R3 → R3 - R1 $(x+1)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & x-2 & 1 \\ 1 & 1 & x-2\end{array}\right|$ So expanding along R1 So |A| = (x + 1)(x - 2)(x - 2) = 0 ⇒ x = -1, 2 for |A| = 0 (singularity) |