Search any question & find its solution
Question:
Answered & Verified by Expert
If rank of $\left(\begin{array}{ccc}x & x & x \\ x & x^2 & x \\ x & x & x+1\end{array}\right)$ is 1, then
Options:
Solution:
1023 Upvotes
Verified Answer
The correct answer is:
$x=0$
$A=\left(\begin{array}{ccc}x & x & x \\ x & x^2 & x \\ x & x & x+1\end{array}\right)$
Since, rank $A=1$, therefore atleast one deteminant of order 1 should be non-zero and all the determinants of order 2 and 3 should be zero.
If $x=0$, then $A=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)$, which have non zero determinant of order 1 only.
If $x=1$, then $A=\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2\end{array}\right)$, which have a non zero determinant of order 2 , nearly $\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right|$
$\therefore x$ can take value 0 only.
Since, rank $A=1$, therefore atleast one deteminant of order 1 should be non-zero and all the determinants of order 2 and 3 should be zero.
If $x=0$, then $A=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)$, which have non zero determinant of order 1 only.
If $x=1$, then $A=\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2\end{array}\right)$, which have a non zero determinant of order 2 , nearly $\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right|$
$\therefore x$ can take value 0 only.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.