If the system of linear equations x + y + z = 2 x + 2y + 3z = 5 2x + 3y + ( ^2 - 1)z = + 1 has infinitely many solutions, then the value of ^2 + ^2 is equal to

Step-by-step solution

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. = vmatrix 1 & 1 & 1 1 & 2 & 3 2 & 3 & ^2 - 1 vmatrix = 0 Expanding along the first row: = 1(2 ^2 - 2 - 9) - 1( ^2 - 1 - 6) + 1(3 - 4) = (2 ^2 - 11) - ( ^2 - 7) - 1 = ^2 - 5 Setting = 0 gives ^2 = 5 . For infinitely many solutions, _z must also be zero. _z = vmatrix 1 & 1 & 2 1 & 2 & 5 2 & 3 & + 1 vmatrix = 0 Expanding along the first row: _z = 1(2 + 2 - 15) - 1( + 1 - 10) + 2(3 - 4) _z = (2 - 13) - ( - 9) - 2 _