linear transformation , inverse?

Question

let L:R^3 map to R^3 be the linear transformation such that
L((x,y,z))= (x, 2x+y, 3x+4y+z)

show that L is bijective. find Inverse of L.

a. find the matrix of L relative the usual basic of R^3
b. find the matric of inverse of L relative the usual basic of R^3

Answer 1

Look at a vector as an 1Xn or nX1 matrix:

Be (x,y,z) a vector represented as a one column matrix

|1 0 0| |x| ... | x |
|2 1 0| |y| = | 2x + y |
|3 4 1| |z| ... | 3x + 4y + z|


b. Now, to find the inverse matrix, Let's put an identity matrix
to its left and perform elementary operations on both

|1 0 0| |1 0 0|
|2 1 0| |0 1 0|
|3 4 1| |0 0 1|

Subtract twice row 1 from row 2

|1 0 0| |1 0 0|
|0 1 0| |-2 1 0|
|3 4 1| | 0 0 1|

subtract thrice row 1 from row 3

|1 0 0| |1 0 0|
|0 1 0| |-2 1 0|
|0 4 1| |-3 0 1|

Subtract 4 times row 2 from row 3

|1 0 0| |1 0 0|
|0 1 0| |-2 1 0|
|0 0 1| |5 -4 1|

Voila

| 1 0 0 |
| -2 1 0 |
| 5 -4 1|

Is L^-1.

Let's check our solution:
Multiply the first row by (x, 2x+y, 3x+4y+z):
1*x + 0*(2x+y) + 0*(3x+4y+z) = x

Multiply the 2nd row by (x, 2x+y, 3x+4y+z):
-2*x + (2x +y ) + 0*(3x+4y+z) = -2x + 2x + y = y

Multiply the 3rd row by (x, 2x+y, 3x+4y+z):
5*x -4*(2x + y) + (3x + 4y + z) = 5x -8x - 4y + 3x + 4y + z = z

Answer 2

You can read off the matrix L:
1 0 0
2 1 0
3 4 1

It's determinant is 1 (clearly non zero since it's a lower triangular matrix with all 1s on the diagonal), the product of the diagonal elements. That show's it's bijective.

The inverse is straightforward, too (if you haven't memorized a way for it, for lower triangular matrices, figure it out row by row (going from top to bottom, and within a row from right to left), by placing L to the right and seeing how you have to fill it in to get the identity matrix when multiplied):
1 0 0
-2 1 0
5 -4 1