Find the matrix B such that AB=t and A=(3 2)
(-1 3)
Hence find the point of intersection of the lines 3x+2y=10 and 3y-4x=x
To find the matrix B, we will solve the equation AB = t.
Given matrix A:
A = [[3, 2],
[-1, 3]]
And target matrix t:
t = [x, y]
Let matrix B be:
B = [[a, b],
[c, d]]
Then, using matrix multiplication:
AB = [[3, 2], [-1, 3]] * [[a, b], [c, d]]
= [[3a + 2c, 3b + 2d], [-a + 3c, -b + 3d]]
Setting this equal to t:
[[3a + 2c, 3b + 2d], [-a + 3c, -b + 3d]] = [x, y]
From the equation above, we can equate the corresponding elements:
Equation 1: 3a + 2c = x
Equation 2: 3b + 2d = y
Equation 3: -a + 3c = 0
Equation 4: -b + 3d = 0
Simplifying equations 3 and 4:
a = 3c
b = 3d
Substituting these values into equations 1 and 2:
3(3c) + 2c = x
3(3d) + 2d = y
Simplifying further:
9c + 2c = x
9d + 2d = y
11c = x
11d = y
Therefore, the matrix B is:
B = [[11c, 11d],
[c, d]]
To find the point of intersection of the lines 3x + 2y = 10 and 3y - 4x = x, we need to solve these two equations simultaneously.
First, let's rearrange the second equation:
3y - 4x = x
3y = 5x
y = (5/3)x
Substituting this expression for y in the first equation:
3x + 2(5/3)x = 10
3x + (10/3)x = 10
(9/3)x + (10/3)x = 10
(19/3)x = 10
19x = 30
x = 30/19
Substituting this value of x into the expression for y:
y = (5/3) * (30/19)
y = 50/19
Therefore, the point of intersection of the lines 3x + 2y = 10 and 3y - 4x = x is (30/19, 50/19).
To find the matrix B, we need to solve the equation AB = t. Since we are given A =
[[3, 2],
[-1, 3]], we can write the equation as follows:
[[3, 2],
[-1, 3]] * B = t
To find B, we can rearrange this equation:
B = (A^-1) * t
First, let's find the inverse of A, denoted as A^-1.
The inverse of A can be found by using the formula:
A^-1 = (1/det(A)) * adj(A)
Where det(A) is the determinant of A, and adj(A) is the adjugate of A.
To find det(A), we calculate the determinant by multiplying the main diagonal elements and subtracting the product of the off-diagonal elements:
det(A) = (3*3) - (2*-1) = 9 + 2 = 11
Next, we find the adjugate of A by swapping the main diagonal elements and changing the sign of the off-diagonal elements:
adj(A) = [[3, -2], [1, 3]]
Now, we can find the inverse of A by multiplying the adjugate of A by 1/det(A):
(1/det(A)) = 1/11
So, A^-1 = (1/11) * [[3, -2], [1, 3]] = [[3/11, -2/11], [1/11, 3/11]]
Now, we can evaluate B by multiplying A^-1 by t.
Given AB = t, we can rearrange it as B = (A^-1)*t.
Let's assume t = [[10], [0]] since we are trying to find the point of intersection.
B = [[3/11, -2/11], [1/11, 3/11]] * [[10], [0]]
By performing the matrix multiplication, we find:
B = [[(3/11)*10 + (-2/11)*0], [(1/11)*10 + (3/11)*0]]
= [[30/11], [10/11]]
Therefore, the matrix B is:
[[30/11], [10/11]]
To find the point of intersection of the lines 3x + 2y = 10 and 3y - 4x = x, we can use the matrix M=[[3, 2], [3, -4]] and the matrix B=[[30/11], [10/11]] to solve the equation Mx = B.
Using matrix multiplication, we have:
[[3, 2], [3, -4]] * x = [[30/11], [10/11]]
Simplifying, we get:
[3x + 2y, 3y - 4x] = [30/11, 10/11]
From the second equation, we have:
3y - 4x = 10/11
Rearranging, we get:
3y = 4x + 10/11
y = (4/3)x + 10/33
Substituting this value of y into the first equation, we have:
3x + 2((4/3)x + 10/33) = 30/11
Multiplying through by 33 to get rid of fractions, we have:
99x + 8x + 80 = 90
107x = 10
x = 10/107
Substituting this value of x into the equation for y, we have:
y = (4/3)(10/107) + 10/33
Simplifying, we get:
y = 40/321 + 10/33
Multiplying through by 321 to get rid of fractions, we have:
321y = 40 + (10 * 321)/33
Simplifying further, we get:
321y = 1320/33 + (10 * 321)/33
321y = (1320 + 10 * 321)/33
y = (1320 + 10 * 321)/(33 * 321)
Simplifying, we have:
y = 2090/10743
Therefore, the point of intersection of the lines 3x + 2y = 10 and 3y - 4x = x is (10/107, 2090/10743).