### Composition and Matrix Multiplication

Exercise For each of the following matrix A, find all the matrices X satisfying AX = I (the identity matrix).

1)

 A = [ 1 2 ]

Answer The matrix X has to be 2 by 1. Let

 X = [ x ] z

Then AX = [x + 2z], and AX = I means x + 2z = 1. Thus we conclude that

 X = [ 1 - 2z ] = [ 1 ] + z[ -2 ] z 0 1

where z is arbitrary.

2)

 A = [ 1 ] 2

Answer The matrix X has to be 1 by 2. Let X = [x y]. Then

 AX = [ x y ] 2x 2y

so that AX = I means x = 1, y = 0, 2x = 0, 2y = 1. The conditions are contradictory. Therefore there is no X such that AX = I.

3)

 A = [ 1 1 ] 1 -1

Answer The matrix X has to be 2 by 2. Let

 X = [ x y ] z w

Then

 AX = [ x + z y + w ] x - z y - w

so that AX = I means

x + z = 1, x - z = 0,
y + w = 0, y - w = 1.

The first two equations tell us x = z = 1/2. The last two equations tell us y = 1/2, w = -1/2. We conclude that AX = I implies

 X = [ 1/2 1/2 ] 1/2 -1/2

4)

 A = [ 2 4 ] -1 -2

Answer The matrix X has to be 2 by 2. Let X be as before. Then

 AX = [ 2x + 4z 2y + 4w ] - x - 2z - y - 2w

so that AX = I means

2x + 4z = 1, - x - 2z = 0,
2y + 4w = 0, - y - 2w = 1.

The first two equations are contradictory. The last two equations are also contradictory. We conclude that there is no X such that AX = I.

5)

 A = [ 0 0 ] 0 0

Answer We have AX = O for all X. Therefore AX can never be the identity matrix.

6)

 A = [ 1 -2 3 ] -2 3 -4

Answer The matrix X has to be 3 by 2. Let

 X = [ x y ] z w u v

Then AX = I means

x - 2z + 3u = 1, - 2x + 3z - 4u = 0,
y - 2w + 3v = 0, - 2y + 3w - 4v = 1.

The first two equations tell us x = 1 + u, z = 2u, u arbitrary. The last two equations tell us y = - 2 + v, w = -1 + 2v, v arbitrary. We conclude that

 X = [ 1 + u -2 +v ] = [ 1 -2 ] + u[ 1 0 ] + v[ 0 1 ] 2u 2v 0 0 2 0 0 2 u v 0 0 1 0 0 1

where u and v are arbitrary.

7)

 A = [ 1 -2 ] -2 3 3 -4

Answer The matrix X has to be 2 by 3. Let

 X = [ x y z ] u v w

Then AX = I means

x - 2u = 1, - 2x + 3u = 0, 3x - 4u = 0,
y - 2v = 0, - 2y + 3v = 1, 3y - 4v = 0,
z - 2w = 0, - 2z + 3w = 0, 3z - 4w = 1.

Note that we actually have three systems of linear equations, with different variables ((x, u) for the first system, (y, v) for the second, (z, w) for the third), but with the same matrix A as the coefficient matrix. Since the system in (x, u) has no solutions, there is no X such that AX = I.