Maths Matrix Worksheet Page 2
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1
2Answer Key
1. The determinant of the matrix is
2p hence we need p = 0.
2. Q Q = I.
1
3. Row-reducing [A I] we find [I A
]. We obtain
8
2 1
1
A
=
3
1
0
.
1
0
0
p
1
p
2
4. Not invertible: the columns are all proportional to the vector
, hence not independent.
. . .
p
5.
(i)
1
0
2
1
1
1
0
1
6
v
=
, v
=
, v
=
.
1
2
3
2
2
1
3
3
3 5
2
2
2
3
2
2
1
(ii) We have u = (v
u)v
+ (v
u) v
+ (v
u) v
= v
+
v
=
.
1
1
2
2
3
3
1
2
3
9
2
10
(iii) V = C(A) where A has as columns the three vectors v
, v
, v
. Then,
1
2
3
6
1
0
2
2
1 0 0
6
2
V
= N (A ) = N
0
1
2 2
= N
0 1 0
2
=
.
2
0
7
5
4
0 0 1
2
1
6. If x
V
and y
V
we show x + y
V . By definition, we have x v = 0 and y v = 0 for all v
V .
Then (x + y) v = x v + y v = 0 for all v
V =
x + y
V . Multiplication by scalars is similar.
7. The matrix is
5
2
4
1
1
A(A A)
A =
2
8
2
.
9
4
2
5
8. Let t = Trace(A). We know from the homework that
2
2
A
tA + (det A)I = 0 =
A
= tA + I.
Assume t = 0. Squaring, we obtain
4
2
2
2
2
2
2
3
2
A
= (A
)
= (tA + I)
= t
A
+ 2tA + I = t
(tA + I) + 2tA + I = A(t
+ 2t) + t
I + I.
4
Since A
= I, we have
2
t
3
2
A(t
+ 2t) =
t
I =
A = uI for u =
.
3
t
+ 2t
3
2
2
The expression u is well defined since the numerator t
+ 2t = t(t
+ 2) = 0. But then det A = u
=
1
which is impossible. Therefore we must have t = 0, and then
2
A
= tA + I = I.
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