Let T: R^3 -> R^3 be a linear transformation whose matrix, with respect to the standard basis is
|1 1 2|
|1 3 0|
|1 0 1|.
If T^(-1){
|96| |u|
| 2| = |v|
| 3| |w|}
then the value of v is?
To find the value of v, we first need to find the inverse of the linear transformation T.
To find the inverse of a linear transformation, we need to find the inverse of its matrix representation. In this case, the matrix representation of T is given as:
|1 1 2|
|1 3 0|
|1 0 1|
To find the inverse of this matrix, we can use matrix operations. One way to do this is by utilizing the concept of elementary row operations to reduce the given matrix to the identity matrix, while performing the same operations on the identity matrix. Let's do these steps together:
Step 1: Create the augmented matrix ([A|I]), where A is the given matrix and I is the identity matrix.
|[1 1 2 | 1 0 0]|
|[1 3 0 | 0 1 0]|
|[1 0 1 | 0 0 1]|
Step 2: Perform elementary row operations to transform the left side of the augmented matrix into the identity matrix.
Step 2.1: Replace R2 with R2 - R1.
|[1 1 2 | 1 0 0]|
|[0 2 -2 | -1 1 0]|
|[1 0 1 | 0 0 1]|
Step 2.2: Replace R3 with R3 - R1.
|[1 1 2 | 1 0 0]|
|[0 2 -2 | -1 1 0]|
|[0 -1 -1 | -1 0 1]|
Step 2.3: Replace R1 with R1 - R3.
|[1 2 3 | 2 0 -1]|
|[0 2 -2 | -1 1 0]|
|[0 -1 -1 | -1 0 1]|
Step 2.4: Replace R2 with R2 + R3.
|[1 2 3 | 2 0 -1]|
|[0 1 -3 | -2 1 1]|
|[0 -1 -1 | -1 0 1]|
Step 2.5: Replace R3 with (-1) * R3.
|[1 2 3 | 2 0 -1]|
|[0 1 -3 | -2 1 1]|
|[0 1 1 | 1 0 -1]|
Step 2.6: Replace R3 with R3 - R2.
|[1 2 3 | 2 0 -1]|
|[0 1 -3 | -2 1 1]|
|[0 0 4 | 3 -1 -2]|
Step 2.7: Replace R3 with (1/4) * R3.
|[1 2 3 | 2 0 -1]|
|[0 1 -3 | -2 1 1]|
|[0 0 1 | 3/4 -1/4 -1/2]|
Step 2.8: Replace R2 with R2 + 3*R3.
|[1 2 3 | 2 0 -1]|
|[0 1 0 | 1/4 -1/4 -1/2]|
|[0 0 1 | 3/4 -1/4 -1/2]|
Step 2.9: Replace R1 with R1 - 3*R3.
|[1 2 0 | -5/4 3/4 5/2]|
|[0 1 0 | 1/4 -1/4 -1/2]|
|[0 0 1 | 3/4 -1/4 -1/2]|
Step 3: The left side of the augmented matrix is now the identity matrix. The right side of the augmented matrix is the inverse matrix ([I|A^(-1)]).
|[1 2 0 | -5/4 3/4 5/2]|
|[0 1 0 | 1/4 -1/4 -1/2]|
|[0 0 1 | 3/4 -1/4 -1/2]|
Therefore, the inverse of the given matrix is:
| -5/4 3/4 5/2 |
| 1/4 -1/4 -1/2 |
| 3/4 -1/4 -1/2 |
Now we can use the given inverse matrix to solve for the column vector [u, v, w] using matrix multiplication:
| -5/4 3/4 5/2 | | 96 |
| 1/4 -1/4 -1/2 | x | 2 |
| 3/4 -1/4 -1/2 | | 3 |
To calculate the value of v, we need to multiply the second row of the inverse matrix with the column vector [96, 2, 3]:
(1/4)(96) + (-1/4)(2) + (-1/2)(3) = 24 - 1/2 - 3/2 = 22
Therefore, the value of v is 22.