Final answer:
To compute the determinant of the matrix, a cofactor expansion is performed resulting in the determinant 6x - 3y + z. To find constants a, b, c for when a vector [x, y, z] lies in the plane P, the cross product of the spanning vectors of the plane is used, resulting in the equation 6x - 3y + z = 0.
Step-by-step explanation:
To compute the determinant of the matrix given by
[ x 1 0 ]
[ y 2 1]
[ z 0 3 ]
we use the rule of Sarrus or the cofactor expansion along the first column (since it contains a zero, which simplifies the process). We'll use the cofactor expansion for this example:
The determinant, denoted as det(A), is calculated by:
det(A) = x * (2*3 - 1*0) - y * (1*3 - 0*0) + z * (1*1 - 2*0)
= x * 6 - y * 3 + z * 1
= 6x - 3y + z
For part (ii), the plane P spanned by vectors [1, 2]T and [0, 1, 3]T can be described by the equation ax + by + cz = 0. Here, the normal vector to the plane is the cross product of the two spanning vectors.
The cross product, denoted as N, is calculated by:
N = [1, 2, 0] x [0, 1, 3]
= (2*3 - 0*1)i - (1*3 - 0*0)j + (1*1 - 2*0)k
= (6)i - (3)j + (1)k
Therefore, the equation of the plane P can be written as:
6x - 3y + 1z = 0
So, the constants a, b, and c are 6, -3, and 1, respectively. This means that a vector [x, y, z]T lies in plane P if and only if 6x - 3y + z = 0.