Quiz 4 Solutions
1. Determine if the given statement is true or false and explain your answer.
(a) If A, B ∈ Mn , then (A − B)2 = A2 − 2AB + B 2 .
This is a false statement. We do not know if the matrices commute, so we can not claim
this is always true. Notice that if we multiply out, we have
(A − B)2 = A2 − AB − BA + B 2
and we have no guarantee that AB = BA.
(b) If T1 (x) and T2 (x) are onto linear transformations from Rn to Rm , then so is W (x) =
T1 (x) + T2 (x).
This is a false statement. Since T1 and T2 are both linear transformations, T1 (0) = 0 and
T2 (0) = 0. But, it is possible that there exists x ∈ Rn such that W (x) = T1 (x)+T2 (x) =
0 but T1 (x) 6= 0 and T2 (x) 6= 0. We could have that T1 (x) = −T2 (x).
(c) If the columns of an n × n matrix A span Rn , then A is singular.
This ia false statement. Since we have n columns in Rn that span the space, then
the columns are linearly independent and therefore we would get a unique solution to
Ax = b for all b ∈ Rn . So, we would have that A is nonsingular.
2. Determine if the given linear transformation is invertible and if so, find T −1 .
x1
4x1 + 3x2
T
=
x2
3x1 + 2x2
First, we write this transformation as a square matrix.
A=
4
3
3
2
Since the determinant is 8 − 9 = −1 6= 0, we have that the matrix is invertible and therefore
the transformation is invertible. To find T −1 , we find A−1 . Using the algorithm, we have
A
−1
1
=
−1
2
-3
-3
4
=
-2
3
3
-4
And so, we have
T −1
x1
−2x1 + 3x2
=
x2
3x1 + −4x2