Exam (elaborations)
MAT1503 EXAM PACK 2024/2025
- Course
- Linear Algebra I (MAT1503)
- Institution
- University Of South Africa (Unisa)
MAT1503 EXAM PACK 2024/2025 QUESTIONS WITH DETAILED ANSWERS. PERFECT FOR EXAM PREPARATION.
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lOMoAR cPSD| 47389193, lOMoAR cPSD| 47389193
1
QUESTION 1
Solution:
1.1).
Area of a triangle with three vertices (a,b) , (c, d) and (e, f) is given by
1a b 1
Area of triangle = |c d 1|
2
e f 1
If the points are collinear then the are of the triangle is equal to zero,then
Area of triangle = 0
1a b 1
|c d 1| = 0
2
e f 1
, lOMoAR cPSD| 47389193
a b 1
|c d 1| = 0 × 2
e f 1
a b 1
|c d 1| = 0
e f 1
We can conclude that the points (a,b) , (c, d) and (e,f) are collinear whenever
a b 1
|c d 1| = 0 e f 1
1.2).
2 −1 3
0 1 1
3[
]
2 1 1 0
2 0 −1 −2
Minors:
1 1 3
M11 = |1 1 0|
0 −1 −2
1 0 1 0 1 1 =1| |−1| |
+3| |
−1 −2 0 −2 0 −1
= 1(−2 − 0) − 1(−2 − 0) + 3(−1 − 0)
= −2 + 2 − 3
= −3
0 1 3
M12 = |2 1 0|
, lOMoAR cPSD| 47389193
1
2 −1 −2
1 0 2 0 2 1
=0| |−1| |+3| |
−1 −2 2 −2 2 −1
= 0 − 1(−4 − 0) + 3(−2 − 2)
= 0 + 4 − 12
= −8
0 1 3
M13 = |2 1 0|
2 0 −2
1 0 2 0 2 1
=0| |−1| |+3| |
0 −2 2 −2 2 0
= 0 − 1(−4 − 0) + 3(0 − 2)
=4−6
= −2
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