math111_logo Inverse


Exercise Find invertible linear transformations and compute the inverse.

1) T1(x1, x2) = (2x1 + 3x2, 5x1 + 7x2): R2R2

Answer The matrix for T1 is

[ 2 3 ]
5 7

Since 2×7 - 3×5 ≠ 0, the matrix is invertible. The inverse is

(2×7 - 3×5)-1[ 7 -3 ] = [ -7 3 ]
-5 2 5 -2

Thus the inverse transformation is T1-1(x1, x2) = (- 7x1 + 3x2, 5x1 - 2x2).

2) T2(x, y) = (2x - 6y, - 5x + 15y): R2R2

Answer The matrix for T2 is

[ 2 -6 ]
-5 15

Since 2×15 - (-5)×(-6) = 0, the matrix is not invertible. Correspondingly, the linear transformation is not invertible.

3) T3(x1, x2, x3) = (- 2x2 + x3, x1 - x3, 3x1 + x2 - 3x3): R3R3

Answer The matrix of the transformation is

[ 0 -2 1 ]
1 0 -1
3 1 -3

We append the identity matrix.

[ 0 -2 1 1 0 0 ]
1 0 -1 0 1 0
3 1 -3 0 0 1

Row operations change this to

[ 1 0 0 1 -5 2 ]
0 1 0 0 -3 1
0 0 1 1 -6 2

Thus we have

[ 0 -2 1 ]-1 = [ 1 -5 2 ]
1 0 -1 0 -3 1
3 1 -3 1 -6 2

and the inverse transformation is T3-1(x1, x2, x3) = (x1- 5x2 + 2x3, - 3x2 + x3, x1 - 6x2 + 2x3).

4) T4(x1, x2, x3) = (- 2x2 + x3, 3x1 + x2 - 3x3): R3R2

Answer The transformation is not invertible because the domain and codomain have different dimensions (see here).

5) T5(x1, x2, x3) = (- 2x2 + x3, x1 - x3, 3x1 + 2x2 - 4x3): R3R3

Answer The matrix of the transformation is

[ 0 -2 1 ]
1 0 -1
3 2 -4

Row operations change this to

[ 1 0 -1 ]
0 -2 1
0 0 0

Thus the matrix, as well as the linear transformation, is not invertible.

6) T6(x1, x2, x3, x4) = (x2 + 2x3 + 4x3, x1 + x2 + 2x3, x1 + x2 + 2x3 - 2x4, x1 + x2 + x3 + 2x4): R4R4

Answer The corresponding matrix is

[ 0 1 2 4 ]
1 1 2 0
1 1 2 -2
1 1 1 2

The matrix is invertible, with

[ -1 3 -2 0 ]
1 -6 4 2
0 2 -1 -1
0 1/2 1/2 0

as the inverse matrix. Thus T6-1(x1, x2, x3, x4) = (- x1 + 3x2 - 2x3, x1 - 6x2 + 4x3 + 2x4, 2x2 - x3 - x4, 1/2 x2 + 1/2 x3).

7) T7(x1, x2, x3, x4, x5) = (x1 + x2, x2 + x3, x3 + x4, x4 + x5, x5 + x1): R5R5

Answer The corresponding matrix is

[ 1 1 0 0 0 ]
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1
1 0 0 0 1

The matrix is invertible, with

1/2[ 1 -1 1 -1 1 ]
1 1 -1 1 -1
-1 1 1 -1 1
1 -1 1 1 -1
-1 1 -1 1 1

as the inverse matrix. Thus T7-1(x1, x2, x3, x4, x5) = 1/2(x1 - x2 + x3 - x4 + x5, x1 + x2 - x3 + x4 - x5, - x1 + x2 + x3 - x4 + x5, x1 - x2 + x3 + x4 - x5, - x1 + x2 - x3 + x4 + x5).

Remark For T(x1, x2, ..., xn) = (x1 + x2, x2 + x3, x3 + x4, ..., xn-1 + xn, xn + x1): RnRn, we have two cases: If n is odd, then T is invertible, with

T-1(x1, x2, ..., xn) =
1/2(x1 - x2 + x3 - ... + xn-2 - xn-1 + xn, x1 + x2 - x3 + ... - xn-2 + xn-1 - xn, - x1 + x2 + x3 - ... + xn-2 - xn-1 + xn, ..., x1 - x2 + x3 - ... + xn-2 + xn-1 - xn, - x1 + x2 - x3 + ... - xn-2 + xn-1 + xn).

If n is even, then for the nonzero vector (1, -1, 1, -1, ..., 1, -1), we have T(1, -1, 1, -1, ..., 1, -1) = 0. Thus T is not one-to-one and is not invertible