0 votes 0 votes Isha Gupta asked Jun 19, 2016 Isha Gupta 1.3k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments ManojK commented Jun 19, 2016 reply Follow Share is it? $\begin{bmatrix} 1 &0 &x \\ 0&x &1 \\ 0 & 1 &x \end{bmatrix}$ and $\begin{bmatrix} x &1 &0\\ x&0&1 \\ 0 & x &1 \end{bmatrix}$ 0 votes 0 votes Isha Gupta commented Jun 19, 2016 reply Follow Share but it has same determinant as mentioned in the question 0 votes 0 votes ManojK commented Jun 19, 2016 i edited by ManojK Jun 19, 2016 reply Follow Share ya its simple . $\begin{vmatrix} 1 &0 &x \\ 0&x &1 \\ 0 & 1 &x \end{vmatrix}$=$x^{2}-1$ $\begin{vmatrix} x &1 &0\\ x&0&1 \\ 0 & x &1 \end{vmatrix}$=$-x^{2}-x$ Given that determinant is same So $x^{2}-1=-x^{2}-x$ $2x^{2}+x-1=0$ Solve the equation you will get $x=-1$ and $x=1/2$ . 2 votes 2 votes Please log in or register to add a comment.
0 votes 0 votes 1 0 x x 1 0 0 x 1 x 0 1 0 1 x 0 x 1 for first matrix take by c1 =1(x2-1)-0+0 =x2=1 ie x=1 for second matrix i took first row =x(0-x)-1(x-0)+0 = -x2-x ie x=-1 it doesn't have same determinant Pravin Paikrao answered Jun 19, 2016 Pravin Paikrao comment Share Follow See all 0 reply Please log in or register to add a comment.