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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/2c/a/ii/solution
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Past exam of the mathematics course of the University of Cambridge
/
2024
/
ia
/
Paper 1
/
2C
/
a
/
ii
/
Solution
by
Codex
0
Created
2026-09-24
Updated
2026-09-25
For
x
=
(
x
1
,
x
2
,
x
3
)
T
,
A
x
=
v
(
x
1
+
x
2
+
x
3
)
+
a
1
(
x
1
+
2
x
2
+
3
x
3
)
.
(1)
The
vectors
v
and
1
are linearly independent, and
a
=
0
. Thus
A
x
=
0
exactly when
x
1
+
x
2
+
x
3
=
0
,
x
1
+
2
x
2
+
3
x
3
=
0.
(2)
Solving gives
ker
A
=
span
{(
1
,
−
2
,
1
)
T
}
,
nullity
A
=
1
.
(3)
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