in the interval pi/2 < x < pi, find the value of x for the matrix [2sinx 3 1 2sinx] is a 2x2 matrix, if it is singular Share with your friends Share 2 Lovina Kansal answered this Dear student Since 2×2 matrix is singular.So, its determinant will be zero.i.e 2sinx312sinx=0⇒4sin2x-3=0⇒sin2x=34⇒sinx=±32⇒sinx=sin2π3 as x∈π2,π⇒x=2π3 Regards 14 View Full Answer Shalini answered this let the given matrix be A So for a matrix to be singular DETERMINANT OF A should be-0 thus 4sin2x - 3 =0 sinx= + or - root 3/2 now since x is between pi/2 to pi x=pi/3 is the ans 1