Given that A=(3 2),find a matrix such that
AB=I -1 3
b)hence determine the point where the lines 3x+2y=10 and 3y-x=4 intersect
To find the matrix B such that AB = I, we need to solve the equation AB = I for B.
The matrix A is given as A = [3 2].
The identity matrix I is given as I = [[-1 3].
We can represent matrix B as B = [a b; c d].
Using matrix multiplication, we have:
AB = [3 2] * [a b; c d]
= [3a + 2c 3b + 2d]
Setting this equal to the identity matrix I, we can equate the corresponding entries:
3a + 2c = -1
3b + 2d = 3
To solve these equations, we can use Gaussian elimination or other methods. Solving these equations, we find:
a = -1
b = 1
c = 1
d = 1
Therefore, the matrix B is B = [-1 1; 1 1].
b) Now, to determine the point where the lines 3x + 2y = 10 and 3y - x = 4 intersect, we can set up a system of equations:
3x + 2y = 10
3y - x = 4
To solve this, we can use matrix multiplication:
[x; y] = B * [10; 4]
= [-1 1; 1 1] * [10; 4]
= [(-1 * 10) + (1 * 4); (1 * 10) + (1 * 4)]
= [-10 + 4; 10 + 4]
= [-6; 14]
Therefore, the point of intersection is x = -6 and y = 14.