Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables tou, v, w that puts qinto the canonical form in `Sylvester's law of inertia'!. What are the values of the associated indices s, t?Select one:O We let u = √3(x − y + 2), v = x+y, w = (x - y)/2 to find that q = u²+². Hence s = 2, t = 0 in Sylvester's law of intertia.O We let u = x - 2y, v = z+y, w = √3y to find that q=u²v² w². Hence s = 1,t = 2 in Sylvester's law of intertia.O we let u = x, v= √2(x+y), w = x+y+z to find that q = u²v² + w²2. Hence s = 2, t = 1 in Sylvester's law of intertia.O None of the others applyOThe quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not aproduct of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.

To find the canonical form of the quadratic form using Sylvester's law of inertia, we need to…

Let q = - 2xy - y² + 2xz+2yz + z² be a quadratic form on R³ viewed as a polynomial in 3 variables. Find a linear change of variables to u, v, w that puts q into the canonical form in `Sylvester's law of inertia'! What are the values of the associated indices s, t? Select one: O We let u = √3(x − y + z), v=x+y, w = (x - y)/2 to find that q = u²+ v². Hence s = 2, t = 0 in Sylvester's law of intertia. w². Hence s = 1, t = 2 in Sylvester's law of intertia. O We let u = x - 2y, v = z+y, w = √3y to find that q =u² - v² ○ we let u = x, v = √√√2(x + y), w =x+y+z to find that q = u² O None of the others apply v² +w². Hence s = 2, t = 1 in Sylvester's law of intertia. O The quadratic form does not obey the condition to be diagonalisable over R. This is because the minimal polynomial of the corresponding matrix is not a product of distinct linear factors. Hence Sylvester's law of inertia does not apply. By convention, we set s = t = ∞ when this happens.