9 votes 9 votes If the two matrices $\begin{bmatrix} 1 &0 &x \\ 0 & x& 1\\ 0 & 1 & x \end{bmatrix}$ and $\begin{bmatrix} x &1 &0 \\ x & 0& 1\\ 0 & x & 1 \end{bmatrix}$ have the same determinant, then the value of $x$ is $\frac{1}{2}$ $\sqrt2$ $\pm \frac{1}{2}$ $\pm \frac{1}{\sqrt2}$ Linear Algebra isro2008 linear-algebra matrix determinant + – jaiganeshcse94 asked May 31, 2016 recategorized Jun 1, 2016 by LeenSharma jaiganeshcse94 2.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 13 votes 13 votes Determinant of $1^{st}$ Matrix $ = x^2 - 1$ Determinant of $2^{nd}$ Matrix $ = -x^2 -x$ Both are equal: $x^2 - 1 = -x^2 - x$ $2x^2 + x -1 =0$ Solve. $x = \frac{1}{2}$ Digvijay Pandey answered Jun 1, 2016 edited Feb 6, 2019 by Digvijay Pandey Digvijay Pandey comment Share Follow See 1 comment See all 1 1 comment reply Anuj Yadav commented Mar 16 reply Follow Share @Sachin Mittal 1x = -1, 1/2but in the options only 1/2 is given that's why we are discarding x=-1. Otherwise both x=-1 and x=1/2 is the answer ? 0 votes 0 votes Please log in or register to add a comment.