Is invertible? det AWS
Determinants 7.1 Introduction to Determinants 2 3 ]. Is 𝐴 invertible? −1 4 det(𝐴) = (2)(4) − (3)(−1) = 11 Since 𝑑𝑒𝑡 ≠ 0, 𝐴 is invertible.
1. Find the determinant of 𝐴 = [
−1 5 ]. Is 𝐵 invertible? 2. Find the determinant of 𝐵 = [ 1/2 0 det(𝐵) = (−1)(0) − (5)(1/2) = 5/2 Since 𝑑𝑒𝑡 ≠ 0, 𝐵 is invertible. 3. Find the determinant of 𝐶 = [
2 −3 ]. Is 𝐶 invertible? 4 −6
det(𝐶 ) = (2)(−6) − (3)(4) = 0 Since 𝑑𝑒𝑡 = 0, 𝐶 is not invertible.
4. For what value(s) of 𝑎 is the matrix 𝐴 = [
2 ] invertible? 0 1 + 𝑎 2𝑎 ] 𝐴=[ −𝑎 −1 So det(𝐴) = (1 + 𝑎)(−1) − (2𝑎)(−𝑎) = −1 − 𝑎 + 2𝑎2 For 𝐴 to be invertible, we need 𝑑𝑒𝑡 ≠ 0. i.e. we need 2𝑎2 − 𝑎 − 1 ≠ 0 (2𝑎 + 1)(𝑎 − 1) ≠ 0 (2𝑎 + 1) ≠ 0 and (𝑎 − 1) ≠ 0 Therefore, 𝑎 ≠ −1/2 and 𝑎 ≠ 1. 𝑎[
1 −1
1 0 ]+ 0 −1
5. Suppose 𝐴 = [
4 −1
1 ] such that 𝑟𝑎𝑛𝑘(𝐴) = 2, find the 3𝑐
value(s) of 𝑐. (hint: use the inverse theorem) If 𝑟𝑎𝑛𝑘(𝐴) = 2, then 𝐴 has full rank. By the inverse theorem, we know all of the following statements are true • 𝐴 is invertible • the RREF of 𝐴 is the identity • det(𝐴) ≠ 0 • 𝐴𝑥⃗ = 𝑏⃗⃗ has a unique solution for all 𝑏⃗⃗ • 𝐴𝑥⃗ = ⃗⃗ 0 has only the trivial solution To find 𝑐, we will use the fact that det(𝐴) ≠ 0 So, (4)(3𝑐) − (1)(−1) ≠ 0 12𝑐 + 1 ≠ 0 Therefore, 𝑐 ≠ −1/12.
7.2 Cofactor Expansion 3 1 0 1. Find the determinant of 𝐴 = [−2 0 1]. 1 5 4 I am choosing to expand along row 1 (there’s a zero) 0 1 −2 1 ] − 1 det [ ]+0 det(𝐴) = +3 det [ 5 4 1 4 = 3(0 − 5) − 1(−8 − 1) + 0 = −6 0 0 3 −1 2. Find thee determinant of 𝐵 = [3 −1 2 0 ] 4 0 1 0 1 0 0 2 I am choosing to expand along column 2 (there are 3 zeros) 0 3 −1 det(𝐵) = −0 + (−1) det [4 1 0 ] − 0 + 0 1 0 2 3 −1 3 −1 ] + 1 det [ ]] − 0 + 0 = 0 − 1 [+0 − 4 det [ 0 2 1 0 = −1[+0 − 4(6 − 0) + 1(0 − (−1))] = −1[0 − 24 + 1] = −[−23] = 23
0 𝑎 1 3. For what value(s) of 𝑎 is the matrix 𝐴 = [4 4 𝑎] 1 𝑎 1 invertible? For the matrix to be invertible, 𝑑𝑒𝑡 ≠ 0. Expanding along row 1: 4 4 4 𝑎 ] + 1 det [ ] det(𝐴) = +0 − 𝑎 𝑑𝑒𝑡 [ 1 1 1 𝑎 = +0 − 𝑎(4 − 𝑎) + 1(4𝑎 − 4) = −4𝑎 + 𝑎2 + 4𝑎 − 4 = 𝑎2 − 4 We need 𝑎2 − 4 ≠ 0 (𝑎 − 2)(𝑎 + 2) ≠ 0 (𝑎 − 2) ≠ 0 and (𝑎 + 2) ≠ 0 Therefore, if 𝑎 ≠ 2 or −2, then 𝐴 is invertible. 3 0 0 4. If 𝐴 = [−1 −1 2 ], find det(𝐴 + 𝐴𝐴−1 ). −3 2 −1 det(𝐴 + 𝐴𝐴−1 ) = det(𝐴 + 𝐼3 ) 3 0 0 1 0 0 = det ([−1 −1 2 ] + [0 1 0]) −3 2 −1 0 0 1 4 0 0 = det [−1 0 2] −3 2 0 Taking the determinant along row 1: 0 2 ] − 0 + 0 = 4(−4) = −16 4 det [ 2 0
7.3 Properties of Determinants 1. Suppose 𝐴 and 𝐵 are 5 × 5 matrices. 𝐵 is obtained by performing the following elementary row operations on 𝐴: i. First multiply row two by 2 ii. Then interchange rows three and four iii. Then multiply every row by −1 iv. Finally, add 3 times row one to row five Find the determinant of 𝐵 in terms of det(𝐴). Interchanging two rows changes the determinant by −1. Multiplying one row by a constant 𝑘 changes the determinant by a multiple of 𝑘. Adding a multiple of one row to another doesn’t change the determinant. i. ii. iii. iv.
𝑑𝑒𝑡 after first ERO is 2det(𝐴) 𝑑𝑒𝑡 after second ERO is −1(2 det(𝐴)) = −2det(𝐴) 𝑑𝑒𝑡 after third ERO is (−1)5 (−2det(𝐴)) = 2det(𝐴) 𝑑𝑒𝑡 after fourth ERO is still 2det(𝐴)
Therefore, det(𝐵) = 2det(𝐴)
2𝑑 2𝑒 2𝑓 𝑎 𝑏 𝑐 2. If 𝐴 = [𝑑 𝑒 𝑓], 𝐵 = [𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖 ], and 𝑔 ℎ 𝑖 𝑔/3 ℎ/3 𝑖/3 det(𝐴) = 3, find det(𝐵). 𝑎 𝑏 𝑐 𝑅 ↔𝑅 𝑑 𝑒 𝑓 2𝑅 2𝑑 2𝑒 2𝑓 1 2 1 [𝑑 𝑒 𝑓 ] → [𝑎 𝑏 𝑐 ] → [ 𝑎 𝑏 𝑐] 𝑔 ℎ 𝑖 𝑔 ℎ 𝑖 𝑔 ℎ 𝑖 1 2𝑑 2𝑒 2𝑓 2𝑑 2𝑒 2𝑓 𝑅 𝑅2 +𝑅3 3 3 [𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖 ] → [𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖 ] → 𝑔 ℎ 𝑖 𝑔/3 ℎ/3 𝑖/3 Using the elementary row operations, 1 1 det(𝐵) = −1(2) ( ) det(𝐴) = −1(2) ( ) 3 = −2 3 3 2 15 7 5 0 10 3 ]. 3. Find the determinant of [0 0 0 0 −4 0 −3 1 0 (hint: use elementary row operations to reduce the matrix into a nicer form) 2 15 7 5 2 15 7 5 0 10 3 ] 𝑅→2 ↔𝑅4 [0 −3 1 0 ] [0 0 0 0 −4 0 0 0 −4 0 −3 1 0 0 0 10 3 2 15 7 5 𝑅3 ↔𝑅4 0 −3 1 0 ] [ → 0 0 10 3 0 0 0 −4 Now the matrix is an upper triangular matrix. 𝑑𝑒𝑡 = (2)(−3)(10)(−4) = 240
4. Given that 𝐼 is the 13 × 13 identity matrix, find det(2𝐼 3 ). det(2𝐼 3 ) = (2)13 det(𝐼 3 ) = 213 det(𝐼 ) = 213. 5. If 𝐴2 = 𝐴, what is det(𝐴)? det(𝐴2 ) = det(𝐴) det(𝐴𝐴) = det(𝐴) det(𝐴) det(𝐴) = det(𝐴) det(𝐴) det(𝐴) − det(𝐴) = 0 det(𝐴) [det(𝐴) − 1] = 0 So, det(𝐴) = 0 or [det(𝐴) − 1] = 0 Therefore, det(𝐴) = 0 or 1
End of Chapter 7 Exam-like Quiz (Determinants) 1 3 −2 1. Given that 𝐴 = [2 0 1 ], find the (3,2)-minor of 𝐴. 4 −1 5 a. 1 b. −1 c. 5 d. −5 e. None of the above Answer c. The (3,2)-minor is 1 −2 ] = 1 − (−4) = 5 𝑀32 = det(𝐴32 ) = det [ 2 1 1 3 −2 2. Given that 𝐴 = [2 0 1 ], find the (3,2)-cofactor of 𝐴. 4 −1 5 a. 1 b. −1 c. 5 d. −5 e. None of the above Answer d. The (3,2)-cofactor is 𝑐32 = (−1)3+2 det(𝐴32 ) = −det [
1 −2 ] = −(1 − (−4)) = −5 2 1
3. If det(𝐴) = 2, det(𝐵) = 3, what is det(𝐴3 𝐵𝑇 𝐴−1 𝐵𝐴𝑇 (𝐵−1 )2 )? a. 2 b. 8 c. −1944 d. −2592 e. None of the above Answer b. det(𝐴3 𝐵𝑇 𝐴−1 𝐵𝐴𝑇 (𝐵−1 )2 ) = det(𝐴3 ) det(𝐵𝑇 ) det(𝐴−1 ) det(𝐵) det(𝐴𝑇 ) det((𝐵−1 )2 ) 1 ) det(𝐵) det(𝐴) det(𝐵−1 )2 = det(𝐴)3 det(𝐵) ( det(𝐴) 2 1 1 = (2)3 (3) ( ) (3)(2) ( ) 2 3
= 23 𝑜𝑟 8
2 1 0 0 3 0 1 ], which of the 4. Suppose that 𝐴 = [ 0 3 5 −1 7 0 1 0 −2 following statements is/are incorrect? i. The rank of 𝐴 is 4 ii. det(𝐴) = 0 iii. The reduced row echelon form of 𝐴 is 𝐼4 (the identity matrix) iv. The homogeneous equation 𝐴x⃗⃗ = ⃗⃗ 0 has a unique solution a. ii only b. i and ii c. iii and iv d. iii only e. They are all correct Answer a. Find the determinant along column 3: 2 1 0 𝑑𝑒𝑡 = +0 − 0 + (−1) det [0 3 1] − 0 0 −2 1 3 1 ] − 0 + 0] = −(2(5)) = −10 ≠ 0 = − [2 det [ −2 1 Since the determinant≠ 0, by the inverse theorem, we know that the matrix has full rank, reduces to the identity, is invertible, and 𝐴x⃗⃗ = ⃗⃗ b and 𝐴x⃗⃗ = ⃗⃗ 0 has a unique solution.
1 0 −3 5. Given that 𝐴 = [7 2 5 ], which of the following 4 1 1 statements is correct? a. The system of linear equations 𝐴𝑥⃗ = 𝑏⃗⃗ always has a unique solution for any vector 𝑏⃗⃗ in ℝ3 b. The RREF of A is the identity matrix c. The homogeneous system 𝐴𝑥⃗ = ⃗0⃗ has only the trivial solution (zero solution) d. 𝐴 is invertible e. None of the above Answer e. 2 det(𝐴) = +1 det [ 1 = −3 − 0 + 3 = 0.
7 5 ] − 0 + (−3) det [ 4 1
2 ] 1
Since the determinant = 0, by the inverse theorem, 𝐴 is not invertible, doesn’t have rank 3, doesn’t reduce to the identity matrix, 𝐴x⃗⃗ = ⃗⃗ b and 𝐴x⃗⃗ = ⃗⃗ 0 don’t have unique solutions.
6. For what value(s) of 𝑎 is the matrix 𝐴 = [ a. 𝑎 ≠ ±10 b. 𝑎 ≠ √2, 𝑎 ≠ √5 c. 𝑎 ≠ ± √2, 𝑎 ≠ ±√5 d. 𝑎 ≠ ±√10 e. 𝐴 is invertible for all values of 𝑎
2 𝑎
𝑎 ] invertible? 5
Answer d. For 𝐴 to be invertible, we need det(𝐴) ≠ 0. So, we need 10 − 𝑎2 ≠ 0 𝑎 ≠ ±√10
7. For what value(s) of 𝑎 is the matrix 𝐴 = [
1 −1
invertible? a. 𝑎 ≠ 7 b. 𝑎 ≠ −7 c. 𝑎 ≠ ±√7 d. 𝑎 ≠ −5 e. 𝐴 is invertible for all real values of 𝑎 Answer e. For 𝐴 to be invertible, we need det(𝐴) ≠ 0. So, we need 𝑎2 + 5 − (−2) ≠ 0 𝑎2 + 7 ≠ 0. Since 𝑎2 + 7 is never 0, 𝐴 is always invertible.
2 ] 𝑎 +5 2
8. Suppose 𝐴 is a 4 × 4 matrix. 𝐴 can be reduced to the 4 × 4 identity matrix 𝐼4 by performing the following elementary row operations: i. First multiply row two by −2 ii. Then add 2 times row one to row four iii. Finally, interchange row one and row three Find the determinant of 𝐴. a. 2 b. −2 c. 1/2 d. −1/2 e. None of the above Answer c. Interchanging two rows changes the determinant by −1. Multiplying one row by a constant 𝑘 changes the determinant by a multiple of 𝑘. Adding a multiple of one row to another doesn’t change the determinant. i. 𝑑𝑒𝑡 after first ERO is −2det(𝐴) ii. 𝑑𝑒𝑡 after second ERO is still −2 det(𝐴) iii. 𝑑𝑒𝑡 after third ERO is −(−2 det(𝐴)) = 2det(𝐴) So, det(𝐼4 ) = 2det(𝐴) 1 = 2det(𝐴) det(𝐴) = 1/2
9. Let 𝐴 be an 𝑛 × 𝑛 matrix, 𝐼𝑛 be the 𝑛 × 𝑛 identity matrix, and 0𝑛 be the 𝑛 × 𝑛 zero matrix. Which of the following statements is always true? a. If 𝐴2 = 𝐴, then 𝐴 = 𝐼𝑛 or 𝐴 = 0𝑛 b. If 𝐴2 = −𝐴, then 𝐴 = 𝐼𝑛 or 𝐴 = 0𝑛 c. If 𝐴2 = 𝐴, then det(𝐴) = 0 or det(𝐴) = 1 d. If 𝐴2 = −𝐴, then det(𝐴) = 0 or det(𝐴) = 1 e. None of the above Answer c. a. Not always true. Counter example 𝐴 = [ b. Not true even if 𝐴 = [
1 0
1 0 ]. 0 1
c. Take the determinant of both sides: det(𝐴2 ) = det(𝐴) det(𝐴) det(𝐴) = det(𝐴) det(𝐴) det(𝐴) − det(𝐴) = 0 det(𝐴) [det(𝐴) − 1] = 0 So, det(𝐴) = 0 or det(𝐴) − 1 = 0. Therefore, det(𝐴) = 0 or 1. d. Not true if 𝐴 = [
1 2
−1 ] −2
−1 ] 0
10. Given that 𝐴 and 𝐵 are both invertible square matrices, which of the following statements is true? a. The scalar multiple of 𝐴 is invertible b. The sum 𝐴 + 𝐵 is invertible c. The difference 𝐴 − 𝐵 is invertible d. The product 𝐴𝐵 is invertible e. None of the above. Answer d. a. False. If we multiply the matrix 𝐴 by 0, it will not be invertible. b. False. Counter example: 1 0 −1 0 1 0 [ ] and [ ] are both invertible, but [ ]+ 0 1 0 −1 0 1 −1 0 0 0 [ ]=[ ] is not invertible. 0 −1 0 0 c. False. Counter example: 1 0 1 0 1 0 0 0 [ ] is invertible, but [ ]−[ ]=[ ] is not 0 1 0 1 0 1 0 0 invertible. d. True. det(𝐴𝐵) = det(𝐴) det(𝐵) but since 𝐴 and 𝐵 are both invertible, neither determinant = 0, so det(𝐴𝐵) ≠ 0. Therefore, 𝐴𝐵 is invertible.
1. Find the determinant of 𝐴 = [
−1 5 ]. Is 𝐵 invertible? 2. Find the determinant of 𝐵 = [ 1/2 0 det(𝐵) = (−1)(0) − (5)(1/2) = 5/2 Since 𝑑𝑒𝑡 ≠ 0, 𝐵 is invertible. 3. Find the determinant of 𝐶 = [
2 −3 ]. Is 𝐶 invertible? 4 −6
det(𝐶 ) = (2)(−6) − (3)(4) = 0 Since 𝑑𝑒𝑡 = 0, 𝐶 is not invertible.
4. For what value(s) of 𝑎 is the matrix 𝐴 = [
2 ] invertible? 0 1 + 𝑎 2𝑎 ] 𝐴=[ −𝑎 −1 So det(𝐴) = (1 + 𝑎)(−1) − (2𝑎)(−𝑎) = −1 − 𝑎 + 2𝑎2 For 𝐴 to be invertible, we need 𝑑𝑒𝑡 ≠ 0. i.e. we need 2𝑎2 − 𝑎 − 1 ≠ 0 (2𝑎 + 1)(𝑎 − 1) ≠ 0 (2𝑎 + 1) ≠ 0 and (𝑎 − 1) ≠ 0 Therefore, 𝑎 ≠ −1/2 and 𝑎 ≠ 1. 𝑎[
1 −1
1 0 ]+ 0 −1
5. Suppose 𝐴 = [
4 −1
1 ] such that 𝑟𝑎𝑛𝑘(𝐴) = 2, find the 3𝑐
value(s) of 𝑐. (hint: use the inverse theorem) If 𝑟𝑎𝑛𝑘(𝐴) = 2, then 𝐴 has full rank. By the inverse theorem, we know all of the following statements are true • 𝐴 is invertible • the RREF of 𝐴 is the identity • det(𝐴) ≠ 0 • 𝐴𝑥⃗ = 𝑏⃗⃗ has a unique solution for all 𝑏⃗⃗ • 𝐴𝑥⃗ = ⃗⃗ 0 has only the trivial solution To find 𝑐, we will use the fact that det(𝐴) ≠ 0 So, (4)(3𝑐) − (1)(−1) ≠ 0 12𝑐 + 1 ≠ 0 Therefore, 𝑐 ≠ −1/12.
7.2 Cofactor Expansion 3 1 0 1. Find the determinant of 𝐴 = [−2 0 1]. 1 5 4 I am choosing to expand along row 1 (there’s a zero) 0 1 −2 1 ] − 1 det [ ]+0 det(𝐴) = +3 det [ 5 4 1 4 = 3(0 − 5) − 1(−8 − 1) + 0 = −6 0 0 3 −1 2. Find thee determinant of 𝐵 = [3 −1 2 0 ] 4 0 1 0 1 0 0 2 I am choosing to expand along column 2 (there are 3 zeros) 0 3 −1 det(𝐵) = −0 + (−1) det [4 1 0 ] − 0 + 0 1 0 2 3 −1 3 −1 ] + 1 det [ ]] − 0 + 0 = 0 − 1 [+0 − 4 det [ 0 2 1 0 = −1[+0 − 4(6 − 0) + 1(0 − (−1))] = −1[0 − 24 + 1] = −[−23] = 23
0 𝑎 1 3. For what value(s) of 𝑎 is the matrix 𝐴 = [4 4 𝑎] 1 𝑎 1 invertible? For the matrix to be invertible, 𝑑𝑒𝑡 ≠ 0. Expanding along row 1: 4 4 4 𝑎 ] + 1 det [ ] det(𝐴) = +0 − 𝑎 𝑑𝑒𝑡 [ 1 1 1 𝑎 = +0 − 𝑎(4 − 𝑎) + 1(4𝑎 − 4) = −4𝑎 + 𝑎2 + 4𝑎 − 4 = 𝑎2 − 4 We need 𝑎2 − 4 ≠ 0 (𝑎 − 2)(𝑎 + 2) ≠ 0 (𝑎 − 2) ≠ 0 and (𝑎 + 2) ≠ 0 Therefore, if 𝑎 ≠ 2 or −2, then 𝐴 is invertible. 3 0 0 4. If 𝐴 = [−1 −1 2 ], find det(𝐴 + 𝐴𝐴−1 ). −3 2 −1 det(𝐴 + 𝐴𝐴−1 ) = det(𝐴 + 𝐼3 ) 3 0 0 1 0 0 = det ([−1 −1 2 ] + [0 1 0]) −3 2 −1 0 0 1 4 0 0 = det [−1 0 2] −3 2 0 Taking the determinant along row 1: 0 2 ] − 0 + 0 = 4(−4) = −16 4 det [ 2 0
7.3 Properties of Determinants 1. Suppose 𝐴 and 𝐵 are 5 × 5 matrices. 𝐵 is obtained by performing the following elementary row operations on 𝐴: i. First multiply row two by 2 ii. Then interchange rows three and four iii. Then multiply every row by −1 iv. Finally, add 3 times row one to row five Find the determinant of 𝐵 in terms of det(𝐴). Interchanging two rows changes the determinant by −1. Multiplying one row by a constant 𝑘 changes the determinant by a multiple of 𝑘. Adding a multiple of one row to another doesn’t change the determinant. i. ii. iii. iv.
𝑑𝑒𝑡 after first ERO is 2det(𝐴) 𝑑𝑒𝑡 after second ERO is −1(2 det(𝐴)) = −2det(𝐴) 𝑑𝑒𝑡 after third ERO is (−1)5 (−2det(𝐴)) = 2det(𝐴) 𝑑𝑒𝑡 after fourth ERO is still 2det(𝐴)
Therefore, det(𝐵) = 2det(𝐴)
2𝑑 2𝑒 2𝑓 𝑎 𝑏 𝑐 2. If 𝐴 = [𝑑 𝑒 𝑓], 𝐵 = [𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖 ], and 𝑔 ℎ 𝑖 𝑔/3 ℎ/3 𝑖/3 det(𝐴) = 3, find det(𝐵). 𝑎 𝑏 𝑐 𝑅 ↔𝑅 𝑑 𝑒 𝑓 2𝑅 2𝑑 2𝑒 2𝑓 1 2 1 [𝑑 𝑒 𝑓 ] → [𝑎 𝑏 𝑐 ] → [ 𝑎 𝑏 𝑐] 𝑔 ℎ 𝑖 𝑔 ℎ 𝑖 𝑔 ℎ 𝑖 1 2𝑑 2𝑒 2𝑓 2𝑑 2𝑒 2𝑓 𝑅 𝑅2 +𝑅3 3 3 [𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖 ] → [𝑎 + 𝑔 𝑏 + ℎ 𝑐 + 𝑖 ] → 𝑔 ℎ 𝑖 𝑔/3 ℎ/3 𝑖/3 Using the elementary row operations, 1 1 det(𝐵) = −1(2) ( ) det(𝐴) = −1(2) ( ) 3 = −2 3 3 2 15 7 5 0 10 3 ]. 3. Find the determinant of [0 0 0 0 −4 0 −3 1 0 (hint: use elementary row operations to reduce the matrix into a nicer form) 2 15 7 5 2 15 7 5 0 10 3 ] 𝑅→2 ↔𝑅4 [0 −3 1 0 ] [0 0 0 0 −4 0 0 0 −4 0 −3 1 0 0 0 10 3 2 15 7 5 𝑅3 ↔𝑅4 0 −3 1 0 ] [ → 0 0 10 3 0 0 0 −4 Now the matrix is an upper triangular matrix. 𝑑𝑒𝑡 = (2)(−3)(10)(−4) = 240
4. Given that 𝐼 is the 13 × 13 identity matrix, find det(2𝐼 3 ). det(2𝐼 3 ) = (2)13 det(𝐼 3 ) = 213 det(𝐼 ) = 213. 5. If 𝐴2 = 𝐴, what is det(𝐴)? det(𝐴2 ) = det(𝐴) det(𝐴𝐴) = det(𝐴) det(𝐴) det(𝐴) = det(𝐴) det(𝐴) det(𝐴) − det(𝐴) = 0 det(𝐴) [det(𝐴) − 1] = 0 So, det(𝐴) = 0 or [det(𝐴) − 1] = 0 Therefore, det(𝐴) = 0 or 1
End of Chapter 7 Exam-like Quiz (Determinants) 1 3 −2 1. Given that 𝐴 = [2 0 1 ], find the (3,2)-minor of 𝐴. 4 −1 5 a. 1 b. −1 c. 5 d. −5 e. None of the above Answer c. The (3,2)-minor is 1 −2 ] = 1 − (−4) = 5 𝑀32 = det(𝐴32 ) = det [ 2 1 1 3 −2 2. Given that 𝐴 = [2 0 1 ], find the (3,2)-cofactor of 𝐴. 4 −1 5 a. 1 b. −1 c. 5 d. −5 e. None of the above Answer d. The (3,2)-cofactor is 𝑐32 = (−1)3+2 det(𝐴32 ) = −det [
1 −2 ] = −(1 − (−4)) = −5 2 1
3. If det(𝐴) = 2, det(𝐵) = 3, what is det(𝐴3 𝐵𝑇 𝐴−1 𝐵𝐴𝑇 (𝐵−1 )2 )? a. 2 b. 8 c. −1944 d. −2592 e. None of the above Answer b. det(𝐴3 𝐵𝑇 𝐴−1 𝐵𝐴𝑇 (𝐵−1 )2 ) = det(𝐴3 ) det(𝐵𝑇 ) det(𝐴−1 ) det(𝐵) det(𝐴𝑇 ) det((𝐵−1 )2 ) 1 ) det(𝐵) det(𝐴) det(𝐵−1 )2 = det(𝐴)3 det(𝐵) ( det(𝐴) 2 1 1 = (2)3 (3) ( ) (3)(2) ( ) 2 3
= 23 𝑜𝑟 8
2 1 0 0 3 0 1 ], which of the 4. Suppose that 𝐴 = [ 0 3 5 −1 7 0 1 0 −2 following statements is/are incorrect? i. The rank of 𝐴 is 4 ii. det(𝐴) = 0 iii. The reduced row echelon form of 𝐴 is 𝐼4 (the identity matrix) iv. The homogeneous equation 𝐴x⃗⃗ = ⃗⃗ 0 has a unique solution a. ii only b. i and ii c. iii and iv d. iii only e. They are all correct Answer a. Find the determinant along column 3: 2 1 0 𝑑𝑒𝑡 = +0 − 0 + (−1) det [0 3 1] − 0 0 −2 1 3 1 ] − 0 + 0] = −(2(5)) = −10 ≠ 0 = − [2 det [ −2 1 Since the determinant≠ 0, by the inverse theorem, we know that the matrix has full rank, reduces to the identity, is invertible, and 𝐴x⃗⃗ = ⃗⃗ b and 𝐴x⃗⃗ = ⃗⃗ 0 has a unique solution.
1 0 −3 5. Given that 𝐴 = [7 2 5 ], which of the following 4 1 1 statements is correct? a. The system of linear equations 𝐴𝑥⃗ = 𝑏⃗⃗ always has a unique solution for any vector 𝑏⃗⃗ in ℝ3 b. The RREF of A is the identity matrix c. The homogeneous system 𝐴𝑥⃗ = ⃗0⃗ has only the trivial solution (zero solution) d. 𝐴 is invertible e. None of the above Answer e. 2 det(𝐴) = +1 det [ 1 = −3 − 0 + 3 = 0.
7 5 ] − 0 + (−3) det [ 4 1
2 ] 1
Since the determinant = 0, by the inverse theorem, 𝐴 is not invertible, doesn’t have rank 3, doesn’t reduce to the identity matrix, 𝐴x⃗⃗ = ⃗⃗ b and 𝐴x⃗⃗ = ⃗⃗ 0 don’t have unique solutions.
6. For what value(s) of 𝑎 is the matrix 𝐴 = [ a. 𝑎 ≠ ±10 b. 𝑎 ≠ √2, 𝑎 ≠ √5 c. 𝑎 ≠ ± √2, 𝑎 ≠ ±√5 d. 𝑎 ≠ ±√10 e. 𝐴 is invertible for all values of 𝑎
2 𝑎
𝑎 ] invertible? 5
Answer d. For 𝐴 to be invertible, we need det(𝐴) ≠ 0. So, we need 10 − 𝑎2 ≠ 0 𝑎 ≠ ±√10
7. For what value(s) of 𝑎 is the matrix 𝐴 = [
1 −1
invertible? a. 𝑎 ≠ 7 b. 𝑎 ≠ −7 c. 𝑎 ≠ ±√7 d. 𝑎 ≠ −5 e. 𝐴 is invertible for all real values of 𝑎 Answer e. For 𝐴 to be invertible, we need det(𝐴) ≠ 0. So, we need 𝑎2 + 5 − (−2) ≠ 0 𝑎2 + 7 ≠ 0. Since 𝑎2 + 7 is never 0, 𝐴 is always invertible.
2 ] 𝑎 +5 2
8. Suppose 𝐴 is a 4 × 4 matrix. 𝐴 can be reduced to the 4 × 4 identity matrix 𝐼4 by performing the following elementary row operations: i. First multiply row two by −2 ii. Then add 2 times row one to row four iii. Finally, interchange row one and row three Find the determinant of 𝐴. a. 2 b. −2 c. 1/2 d. −1/2 e. None of the above Answer c. Interchanging two rows changes the determinant by −1. Multiplying one row by a constant 𝑘 changes the determinant by a multiple of 𝑘. Adding a multiple of one row to another doesn’t change the determinant. i. 𝑑𝑒𝑡 after first ERO is −2det(𝐴) ii. 𝑑𝑒𝑡 after second ERO is still −2 det(𝐴) iii. 𝑑𝑒𝑡 after third ERO is −(−2 det(𝐴)) = 2det(𝐴) So, det(𝐼4 ) = 2det(𝐴) 1 = 2det(𝐴) det(𝐴) = 1/2
9. Let 𝐴 be an 𝑛 × 𝑛 matrix, 𝐼𝑛 be the 𝑛 × 𝑛 identity matrix, and 0𝑛 be the 𝑛 × 𝑛 zero matrix. Which of the following statements is always true? a. If 𝐴2 = 𝐴, then 𝐴 = 𝐼𝑛 or 𝐴 = 0𝑛 b. If 𝐴2 = −𝐴, then 𝐴 = 𝐼𝑛 or 𝐴 = 0𝑛 c. If 𝐴2 = 𝐴, then det(𝐴) = 0 or det(𝐴) = 1 d. If 𝐴2 = −𝐴, then det(𝐴) = 0 or det(𝐴) = 1 e. None of the above Answer c. a. Not always true. Counter example 𝐴 = [ b. Not true even if 𝐴 = [
1 0
1 0 ]. 0 1
c. Take the determinant of both sides: det(𝐴2 ) = det(𝐴) det(𝐴) det(𝐴) = det(𝐴) det(𝐴) det(𝐴) − det(𝐴) = 0 det(𝐴) [det(𝐴) − 1] = 0 So, det(𝐴) = 0 or det(𝐴) − 1 = 0. Therefore, det(𝐴) = 0 or 1. d. Not true if 𝐴 = [
1 2
−1 ] −2
−1 ] 0
10. Given that 𝐴 and 𝐵 are both invertible square matrices, which of the following statements is true? a. The scalar multiple of 𝐴 is invertible b. The sum 𝐴 + 𝐵 is invertible c. The difference 𝐴 − 𝐵 is invertible d. The product 𝐴𝐵 is invertible e. None of the above. Answer d. a. False. If we multiply the matrix 𝐴 by 0, it will not be invertible. b. False. Counter example: 1 0 −1 0 1 0 [ ] and [ ] are both invertible, but [ ]+ 0 1 0 −1 0 1 −1 0 0 0 [ ]=[ ] is not invertible. 0 −1 0 0 c. False. Counter example: 1 0 1 0 1 0 0 0 [ ] is invertible, but [ ]−[ ]=[ ] is not 0 1 0 1 0 1 0 0 invertible. d. True. det(𝐴𝐵) = det(𝐴) det(𝐵) but since 𝐴 and 𝐵 are both invertible, neither determinant = 0, so det(𝐴𝐵) ≠ 0. Therefore, 𝐴𝐵 is invertible.
