Let A be the matrix below and define a transformation T:ℝ3→ℝ3 by T(U) = AU. For the vector B below, find a vector U such that T maps U to B, if possible. Otherwise state that there is no such U.
A =
+3+3 −6
+1+3 −8
−3 0 −2
B =
+6
0
−9
Not sure how to start this question..please help!
U =
2
-1
3
To find a vector U such that T maps U to B, we need to solve the equation T(U) = B.
First, we need to understand how the transformation T works. It is defined as T(U) = AU, where A is the given matrix. This means that we multiply the matrix A by the vector U to obtain the transformed vector T(U).
Given matrix A:
A =
[3 3 -6]
[1 3 -8]
[-3 0 -2]
Given vector B:
B =
[6]
[0]
[-9]
To find U, we need to solve the equation AU = B. This can be done by multiplying the inverse of matrix A to both sides of the equation.
To find the inverse of matrix A, we need to check if A is invertible (i.e., if the determinant of A is non-zero).
To calculate the determinant of A, we use the formula:
det(A) = (3 * 3 * (-2) + 3 * (-8) * 0 + (-6) * 1 * 0) - ((-6) * 3 * (-3) + 3 * 1 * (-2) + 0 * 3 * 0)
= (-18 - 0 + 0) - (-54 + 6 + 0)
= -18 + 60
= 42
Since the determinant of A is non-zero (42 ≠ 0), matrix A is invertible.
Therefore, we can find the inverse of A and solve the equation AU = B.
To find the inverse of A, we calculate the adjugate matrix (adj(A)) and multiply it by the reciprocal of the determinant of A.
The adjugate matrix of A is obtained by taking the transpose of the cofactor matrix of A.
The cofactor matrix of A is obtained by calculating the determinant of each of the 3x3 submatrices of A, multiplied by (-1) raised to the power of the sum of row number and column number of each element.
Once we have the adjugate matrix, we multiply it by the reciprocal of the determinant of A to get the inverse of A.
However, before proceeding with the calculations, we need to confirm that A is a square matrix (number of rows = number of columns).
In this case, A is a 3x3 matrix, so the calculations can proceed.
First, we calculate the cofactor matrix of A:
Cofactor (C) =
[(-1)^(1+1) * det([3 -8; 0 -2]) (-1)^(1+2) * det([1 -8; -3 -2]) (-1)^(1+3) * det([1 3; -3 0])]
[(-1)^(2+1) * det([-3 -6; 0 -2]) (-1)^(2+2) * det([3 -6; -3 -2]) (-1)^(2+3) * det([3 3; -3 0]))]
[(-1)^(3+1) * det([-3 -6; 1 3]) (-1)^(3+2) * det([3 -6; 1 3]) (-1)^(3+3) * det([3 3; 1 3]))]
Next, we find the adjugate matrix of A by taking the transpose of the cofactor matrix:
adj(A) =
[(-1)^(1+1) * det([3 -8; 0 -2]) (-1)^(2+1) * det([3 -6; 0 -2]) (-1)^(3+1) * det([-3 0; -3 -2]) ]
[(-1)^(1+2) * det([1 -8; -3 -2]) (-1)^(2+2) * det([-3 -6; -3 -2]) (-1)^(3+2) * det([3 3; -3 0]) ]
[(-1)^(1+3) * det([1 3; -3 0]) (-1)^(2+3) * det([3 3; -3 0]) (-1)^(3+3) * det([3 3; 1 3]) ]
Next, we calculate the determinant of A:
det(A) = 42
Finally, we calculate the inverse of A by multiplying the adjugate matrix by the reciprocal of the determinant:
A^(-1) = (1 / 42) * adj(A)
Now that we have obtained the inverse of A, we can find U by multiplying the inverse of A to B:
U = A^(-1)B
Substitute the values of A^(-1) and B to find U. If the solution exists, it will be the vector U such that T(U) = B.
To find a vector U such that T maps U to B, we need to solve the equation AU = B.
Given matrix A:
A = [3 3 -6; 1 3 -8; -3 0 -2]
Given vector B:
B = [6; 0; -9]
Let's set up the equation AU = B and solve for vector U:
[3 3 -6; 1 3 -8; -3 0 -2] * [x; y; z] = [6; 0; -9]
This can be represented as a system of linear equations:
3x + 3y - 6z = 6
x + 3y - 8z = 0
-3x - 2z = -9
We can solve this system of equations using various methods such as substitution, elimination, or matrix methods. Let's use the matrix method here.
We can rewrite the system of equations in matrix form as follows:
[3 3 -6; 1 3 -8; -3 0 -2] * [x; y; z] = [6; 0; -9]
Let's augment the matrix [A | B]:
[3 3 -6 | 6]
[1 3 -8 | 0]
[-3 0 -2 | -9]
Applying row operations to obtain an upper triangular matrix:
R2 = R2 - (1/3) * R1
R3 = R3 + R1
[3 3 -6 | 6]
[0 2 -6 | -2]
[0 3 -8 | -3]
R3 = R3 - (3/2) * R2
[3 3 -6 | 6]
[0 2 -6 | -2]
[0 0 1 | 0]
From the last row, we get z = 0.
Substituting this value back into the second row, we get:
2y - 6(0) = -2
2y = -2
y = -1
Substituting the values of y and z into the first row, we get:
3x + 3(-1) - 6(0) = 6
3x - 3 = 6
3x = 9
x = 3
Therefore, the vector U that maps to B under the transformation T is:
U = [3; -1; 0]
Thus, U is [3; -1; 0].