If ( [ array ccc 1 & -1 & x 1 & x & 1 x & -1 & 1 array ] ) has no inverse, then the real value of (x ) is

Step-by-step solution

It is given that matrix (A= [ array ccc 1 & -1 & x 1 & x & 1 x & -1 & 1 array ] ) have no inverse, so ( array rlrl & |A|=0 | array ccc 1 & -1 & x 1 & x & 1 x & -1 & 1 array | & =0 1(x+1)+1(1-x)+x (-1-x^2 ) & =0 x+1+1-x-x-x^3 & =0 x^3+x-2=0 (x-1) (x^2+x+2 ) & =0 array ) Either (x=1 ) or (x^2+x+2=0 ) But discriminant of quadratic equation (x^2+x+2=0 ) is negative so no real roots. ( |A|=0 x=1 ) Hence, option (d) is correct