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Summary mat1503 exam pack
- Course
- Linear Algebra I (MAT1503)
- Institution
- University Of South Africa (Unisa)
Summary of 30 pages for the course Linear Algebra I at Unisa (mat1503 exam pack)
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MAT1503Assignment 9 2021
Memorandum
QUESTION 1
Consider the system of linear equations, where the information regarding the system is
represented by the augmented matrix below:
k 1 1 0
E = [0 k − 1 0 | −2 ]
0 0 k2 − 1 k + 1
The system is inconsistent for:
a. k=0
b. k = 2 and k = −1
c. k=1
d. k = −1
SOLUTION
The system is already in REF. To determine the solutions, we look at the last row.
No solution Exactly one solution Infinitely many solutions
For the above system to be For the above system to be For the above system to have
inconsistent (no solution), the consistent (exactly one infinitely many solutions the last
last row must be of the form: solution), the last row must be row must be of the form:
of the form:
[0 0 0| ∗] [0 0 0| 0]
[0 0 ∗| ∗]
where * is any real number Thus, k2 − 1 = 0 and k + 1 = 0
except zero (0). where * are any real numbers ⇒ k = ±1 . Thus,
except zero (0). for the above system to have
Thus, k2 − 1 = 0 and k + 1 ≠ 0 infinitely many solutions
⇒ k = ±1 and k ≠ −1. Thus, Thus, k2 − 1 ≠ 0 and k + 1 ≠ 0 𝐤 = −𝟏.
for the above system to be ⇒ k ≠ ±1 . Thus,
inconsistent 𝐤 = 𝟏 for the above system to have
exactly one solution
𝐤 ∈ ℝ, except 𝐤 ± 𝟏.
©2021
JTS Maths Tutoring
Keeping you mathematically informed
Page 1 of 30
, MAT1503
Ass 9 2021
Memorandum
QUESTION 2
Given two square matrices X and Y of the same size such that (X − Y)(Y + X) = X2 − Y2.
Which of the following statements is correct?
a. XY = O, where O is the zero matrix of the same dimension as X and Y.
b. X=Y
c. X and Y are inverses of each other
d. XY = YX
SOLUTION
(X − Y)(Y + X) = X2 − Y2 XY = YX in those special cases
where X and Y commute ∵ matrix
(X − Y)(Y + X) = X2 − Y2 multiplicand is not commutative
in general.
X. Y + X. X − Y. Y − Y. X = X2 − Y2
XY + X2 − Y2 − YX = X2 − Y2
X2 − Y2 + XY − YX = X2 − Y2
Now for the LHS to be equal to the RHS ⇒ XY = YX, hence XY −
YX = zero matrix.
©2021
JTS Maths Tutoring
Keeping you mathematically informed
Page 2 of 30
, MAT1503
Ass 9 2021
Memorandum
QUESTION 3
Given that Z = XY, where X and Y are the matrices given below:
−1 2
X = [ 3 −2 −1 2 −4 1
] and Y = [ 1 0 ]
5 0 7 2
4 1
Now find the size of Z and add together the following entries Z23, Z32, Z34, Z43 and Z14 to
get the resulting sum:
a. Z is a 4 x 4 matrix and the sum is equal to − 100
b. Z is a 4 x 4 matrix and the sum is equal to 47
c. Z is a 4 x 4 matrix and the sum is equal to − 17
d. Z is a 4 x 4 matrix and the sum is equal to100
SOLUTION
−1 2 −1. −1 + 2.1 −1.2 + 2.0 −1. −4 + 2.7 −1.1 + 2.2
3 −2 −1 2 −4 1] = 3. −1 + −2.1 3.2 + −2.0 3. −4 + −2.7 3.1 + −2.2
Z = XY = [ ][ ]
5 0 1 0 7 2 [ 5.2 + 0.0 5. −4 + 0.7 5.1 + 0.2
5. −1 + 0.1
4 1 4. −1 + 1.1 4.2 + 1.0 4. −4 + 1,7 4.1 + 1.2
3 −2 18 3
−5 6 −26 1
=[ ]
−5 10 −20 5
−3 8 −9 6
Z23 = −26, Z32 = 10, Z34 = 5, Z43 = −9 and Z14 = 3 Z23 refers to the entry at row 2 and
column 3 in the given matrix Z.
Z23 + Z32 + Z34 + Z43 + Z14 = −26 + 10 + 5 − 9 + 3 = −17
©2021
JTS Maths Tutoring
Keeping you mathematically informed
Page 3 of 30
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