MTH007 LINEAR ALGEBRA
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2022 Sample Paper Solution
BACHELOR DEGREE - Year 1
MTH007
LINEAR ALGEBRA
Part I: True or False (3 marks for each question, 30 marks in total).
Question 1. If the matrices A and B satisfying AB = BA, then A and B must be square matrix of same size.
Question 2. If A and B are two nonsingular matrices of the same size, then A + B must be nonsingular and (A + B) − 1 = A − 1 + B − 1 .
Question 3. Let v1 , v2 and v3 are linearly dependent vectors, then v1 must be expressible as a linear combination
of v2 and v3 .
Question 4. The solution set of a non-homogeneous system Ax = b can be expressed as the sum of a particular solution of this system Ax = b and the general solution of Ax = 0.
Question 5. Let A be an n × n, n > 3 matrix, then det(2A) = 2det(A).
Question 6. The set of vectors consisting of v1 , v2 , v3 and v4 is linearly independent, where
┌ 1┐ ┌ 2┐ ┌ 1┐ ┌ 1┐
v1 = ''1(1)'' , v2 = ''0(1)'' , v3 = ''0(2)'' , v4 = ''0(1)'' .
Question 7. If B is obtained from a matrix A by several elementary row operations, then rank(B) = rank(A).
Question 8. The dimension of the column space of a matrix A equals to the dimension of its null space .
Question 9. Let A be an m × n matrix. If there exists an n × m matrix C such that CA = I纮 , then m > n.
Question 10. If λ is an eigenvalue of the square matrix A, then λ2 + 2 must be the eigenvalue of A2 + 2I .
Part II: Multiple Choice questions (3 marks for each question, 30 marks in total)
Question 11.Which of the following matrices is a reduced echelon matrix? Answer
┌ 1 0 2 0 ┐ ┌ 1 1 2 0 ┐ ┌ 1 0 2 0 ┐
A . ' 0 1 1 0 '; B . ' 0 1 3 0 '; C . ' 0 1 - 1 1 ';
' 0 0 -2 1 ' ' 0 0 -2 1 ' ' 0 0 -2 1 '
┌ 0 1 1 1 ┐ ┌ 1 0 2 0 ┐
D. ' -1 1 -1 1 '; E. ' 0 1 1 0 '
Question 12. Which of the following homogeneous system of linear equations has nontrivial solutions ? Answer
| 3x1 + 5x2 - 4x3 = 0 | 3x1 + 5x2 - 4x3 = 0 | 3x1 + 5x2 - 4x3 = 0 | 3x1 + 5x2 - 4x3 = 0
| | | |
Question 13. Let
┌ 1 ┐ ┌ 5 ┐ ┌ -3 ┐ ┌ -4 ┐
v1 = '' -(-)2(1) '' , v2 = '' -(-)7(4) '' , v3 = '' 0(1) '' and ϕ = '' α(3) '' .
For what value of α will the vector ϕ be in Span{v1 , v2 , v3 } ?
Answer
A. 1; B. 5; C. 10; D. 0. E. None of the answers above.
Question 14. Given the n × n matrix A, B and C, Which of the following statement is NOT always true ? Answer
A. A + B = B + A;
B. (A + B)T = AT + BT ;
C. A(B + C) = AB + AC;
D. AB = BA;
E. None of the answers above, i.e. all the answers above are correct.
Question 15. Let A and B be nonsingular square matrices of the same size. Which of the following statement is NOT always true ? Answer
A. det((AB)T ) = det(AB);
B. det(AB) = det(A)det(B);
C. det(A + B) = det(A) + det(B);
D. det((AB) − 1 ) = ;
E. None of the answers above, i. e. all the statements above are true.
Question 16. Suppose
│ │ = 7.
Which of the following determinant is not 7? Answer
A. │ a d b e c f │; B. 6(1) │ 2 d 2b e 2c f │; C. │ │;
D. │ a 2d b 2e c 2f │; E. │ a 2d b 2e c 2f │ .
Question 17 Which of the following statement is NOT always true ? Answer
A. If A is an m × n matrix and the equation Ax = b is consistent for some b, then the columns of A span Rm .
B. If linear system Ax = b has distinct solutions, then so does the linear system Ax = 0.
C. If A is an m × n matrix, and Ax = b has distinct solutions. If Ax = c is consistent, then the linear system Ax = c has distinct solutions.
D. If A is a nonsingular matrix, then columns of A are linearly independent.
E. None of the answers above, i.e. all the statements above are true.
Question 18. Which of the following statement is NOT always true ? Answer
A. If A is a square matrix and the equation Ax = 0 has only the trivial solution, then A is row equivalent to the n × n identity matrix.
B. If the equation Ax = b has more than one solution, then Ax = 0 has infinite many solutions.
C. If A is an n × n matrix, then the equation Ax = b has at least one solution for each b e R纮 .
D. If AT is not invertible, then A must be not invertible.
E. If A and B are n × n nonsingular matrices, then AB is invertible.
Question 19. Determine the value of c such that the vectors v1 , v2 and v3 are linearly dependent, where
┌ c┐ ┌ 1┐ ┌ 1┐
v1 = ''1(1)'' , v2 = ''1(c)'' , v3 = ''c(1)'' .
Answer
A. c = 1
B. c = -2;
C. c = 1 or c = -2;
D. c = -1 or c = 2;
E. None of the answers above
Question 20.
Find all the values of a and b such that the following linear system has no solution?
| ax1 + x2 + x3 = 4
x1 + bx2 + x3 = 3
│ x1 + 2bx2 + x3 = 4
Answer
A. a = 1
B. b = 0;
C. a = 1 or b = 0;
D. a = 1 and b , or b = 0;
E. None of the answers above
Part III: Blank filling questions (3 marks for each question, 18 marks in total)
Question 21. Given the matrix A = [1(4) 0
l 1 」 l 1 」
Question 22. Let u = '「 3(2) l(') and v = '「 4-1 l(') . Then (uvT )10
经
,
Question 23. Solve the linear system and find the solution〈
,
(
-1」
2(1) l('), then A + BT = .
Answer:
= .
Answer: (uvT )10 = 69
2x1 - x2 - x3 = 4
3x1 + 4x2 - 2x3 = 11 ; 3x1 - 2x2 + 4x3 = 11
A = [0(5)
1 ' 2 |
4 8 12 |
-1 」 -2 ' -3 l . |
= .
1
Question 24 . Given A = ' 1
「3
l 3 」
Answer x = ' 1 '
-1 1」
1 0l('), find the inverse of the matrix A, A − 1 = .
l 1 3 -1」
Answer: A − 1 = '-1 -2 1 '
Question 25. Consider the matrix 3 ' 1 A = ' ' 1 「 |
1 3 1 1 |
1 1 3 1 |
1 」 1 ' ' . 1 ' l |
Then the determinant of matrix A equals to .
Question 26. Find the dimension of the vector space spanned by the vectors v1 , v2 and v3 ,
l 1 」 l 9 」 l-2」
' 2 ' ' 100 ' ' -4 '
where v1 = ' ', v2 = ' ', v3 = ' '.
'- 1 ' ' 10 ' ' 2 '
「 l 「 l 「 l
Answer: 48.
Answer .
Answer 2.
Part IV: Comprehensive Questions (11 marks for each question, 22 marks in total)
Please write down your solutions with detailed justifications.
Question 27. (11 marks)
Consider the matrix
┌ 1 3 -2 1 ┐
A = ' 2 1 3 2 '
' 3 4 5 6 ' .
1. Find the reduced echelon form for the matrix A.
2. Find a basis for N(A) (the null space of A) and the dimension of N(A).
3. Find the dimension of the column space of A and the row space of A.
4. Give a reason why the dimension of the row space equals to the dimension of the column space.
Solution:
1. By performing elementary row operations, the reduced echelon form can be found as follows:
┌ 1 3 -2 1 ┐ ┌ 1 3 -2 1 ┐ ┌ 1 3 -2 1 ┐ ┌ 1 0 0 -13/20 ┐
A = ' 2 1 3 2 ' ~ ' 0 -5 7 0 ' ~ ' 0 -5 7 0 ' ~ ' 0 1 0 21/20 '
┌ 13 ┐
' -21 '
2. Therefore a basis for the null space is ' ' and the dimension of the null space is dimN(A) = 1 .
' - 15 '
' 20 '
3. Based on the reduced echelon form of A, rank(A) = 3 and the first, second and the third columns of A are linearly independent. (or the first second and fourth columns) Therefore,
┌ 1 ┐
v1 = ' 2 '
┌ 3 ┐
v2 = ' 1 '
' 4 '
┌ -2 ┐
and v3 = ' 3 '
' 5 '
form a basis for column space of A and further the dimension of column space is 3. . The row vectors
| 1 0 0 -13/20 ]T , | 0 1 0 21/20 ]T , | 0 0 1 3/4 ]T
form a basis for the row space since elementary row operation does not change the row space and the dimension of the row space is 3.
4. Row reduction does not change the row space and does not change the linearly independency of column vectors and therefore the dimension of row space equals to the number of nonzero rows in the reduced echelon form and equals to the number of pivot elements, which equals to the dimension of column space.
Question 28. (11 marks)
┌ 3 ┐ ┌ -1 ┐ ┌ -2 ┐
Given the vectors v1 = '' 2 '' , v2 = '' 0 '', and v3 = '' -2 '' and let A = [v1 v2 v3].
1. Prove that the two vectors v1 and v2 are linearly independent.
2. Prove that the matrix A is singular (not invertible).
3. Find all the eigenvalues and the corresponding eigenvectors for each eigenvalue.
4. Prove that the any two eigenvectors belonging to distinct eigenvalues are linearly independent for a square matrix A.
5. Prove that the any three eigenvectors belonging to distinct eigenvalues are linearly independent for a square matrix A.
Solution:
1. (method 1) Two vectors are linearly dependent if and only if one vector is the multiple of the other one. But from this questions we can’t get v1 = λv2 or v2 = λv1 (easy to see that the component of v1 and v2
are not proportional). Therefore these two vectors are linearly independent.
(method 2)
┌ 3 -1 ┐ ┌ 1 0 ┐ ┌ 1 0 ┐
A = ' 2 0 ' ~ ' 2 0 ' ~ ' 0 1 '
' 2 -1 ' ' 2 -1 ' ' 0 0 '
Thus v1 and v2 are linearly independent as the system c1 v1 + c2 v2 = 0 has only the trivial solution.
2. (Method 1)
┌ 3 -1 2 ┐ ┌ 1 0
A = ' 2 0 -2 ' ~ ' 2 0
' 2 -1 -1 ' ' 2 -1
-1 ┐ ┌ 1
-2 ' ~ ' 0
-1 ' ' 0
0 -1 ┐
0 1 '
0 0 '
Therefore A is singular.
(Method 2) By calculating of the determinant of matrix A, we can get det(A) = 0 and therefore A is singular.
3.
det(A - λI) = │ 3 -22 λ ---122- λ │ = -λ(λ - 1)2 .
Therefore the eigenvalues are 0, 1 and 1. Solving the system Ax = 0, we can get the eigenvectors for the eigenvalue 0 as follows: t1 [111]T , t1 e R and t1 0.
Solving the system (A - I)x = 0, we can get the eigenvectors for the eigenvalue 1 as follows: t2 [120]T + t3 [0 - 21]T , t2 and t3 e R and t2(2) + t3(3) 0.
4. Suppose wi is an eigenvector of the matrix A for the eigenvalue λi , for i = 1, 2, and λ 1 , λ2 are the distinct eigenvalues λ 1 λ2 . Then Aw1 = λ 1 w1 and Aw2 = λ2 w2 . Next we will use the proof by contradiction. If w1 and w2 are linearly dependent, then w2 = γw1 and therefore λ2 w2 = Aw2 = A(γw1 ) = γAw1 = γλ 1 w1 and λ2 w2 = λ2 γw1 = γλ2 w1 . Therefore γ(λ2 - λ 1 )w1 = 0. As λ2 λ 1 and w1 0, thus γ = 0. and further w2 = γw1 = 0, which is contradicted with w2 is an eigenvector of A. Therefore the vectors w 1 and w 2 are linearly independent.
5. Suppose wi is an eigenvector of the matrix A for the eigenvalue λi , for i = 1, 2, 3, and λ 1 , λ2 and λ3 are the distinct eigenvalues. Next we need to prove that w1 , w2 and w3 are linearly independent. We would use the proof by contradiction. Suppose that the vectors w1 , w2 and w3 are linearly dependent. Then there must be one vector, without loss of generality we assume that w3 can be linearly expressed by w1 and w2 . i.e. there exists c1 and c2 such that w3 = c1 w1 + c2 w2 . Actually from part 4 we have known that w1 and w2 are linearly independent as they are the eigenvectors belonging to λ 1 and λ2 satisfying λ 1 λ2 . multiplying λ3 for both sides of the equality w3 = c1 w1 + c2 w2 , we can get λ3 w3 = c1 λ3 w1 + c2 λ3 w2 . Therefore c1 (λ1 - λ3 )w1 + c2 (λ2 - λ3 ) = 0. By the linearly independency of w1 and w2 , λ 1 λ3 , and λ2 λ3 , c1 = c2 = 0 and further w3 = 0, which is contradicted with w3 is an eigenvector. Therefore the vector set w1 , w2 and w3 are linearly independent.
2023-01-04