Last name:
name:
1
Quiz 2 (Notes, books, and calculators are not authorized).
Show all your work in the blank space you are given on the exam sheet. Always justify your
answer. Answers with no justification will not be graded.
−1 0 1
1 3 5
Question 1: Let A =
,B=
. Find (a) A + B (b) 2A − 3B (c) BAT .
1 2 1
2 4 6
0 3 6
A+B =
3 6 7
−5 −9 −12
2A − 3B =
−4 −8 −16
4 12
T
BA =
4 16
Question 2: Find the inverse of A =
5
6
6
.
7
The determinant of A is |A| = 35 − 36 = −1 6= 0. The matrix is invertible. We use the formula
from class to compute the inverse:
1
7 −6
−7 6
−1
A =
=
6 −5
−1 −6 5
2
Quiz 2, September 16, 2014
T
Question 3: A complex square matrix M is said
which
√ if M M = I. Determine
to be unitary
1
+
i
−
i
1
−
i
3
of the following matrices are unitary: A = √15 √
.
, B = √13
1−i
i
3 1−i
T
We compute AA ,
1 1+i
√
AA =
3
5
T
√ − 3 1−
√i
1−i − 3
√ 3
1
=
0
1+i
0
.
1
T
This proves that A is unitary. We compute BB ,
1
1
T
i
1−i
3
−i 1 + i
BB =
=
i
1 + i −i
3 1−i
3 −i − 1 + i − 1
1 3
i−1−i−1
=
3
3 −2
−2
3
This proves that B is not unitary.
Question 4: Find a 2×2 real-valued matrix P whose columns are unit orthogonal vectors and
whose second column is a multiple of (1, −2)T .
The second column of P is λ(1, −2) where λ ∈ R. The norm of this vector is λ2 (1 + 4) = 5λ2 .
Since this is a unit vector, we infer that λ = ± √15 . Let us choose λ = √15 . The first column of P
is (a, b)T , i.e.,
"
#
√1
a
5
P =
b − √25
with a2 + b2 = 1 and a √15 − b √25 = 0. This means a = 2b and b2 (1 + 4) = 1, i.e., a = ± √25 and
b = ± √15 . In conclusion we can choose
1 2
P =√
5 1
1
.
−2