System of Linear Equations 5.1 Introduction to Systems of Linear
System of Linear Equations 5.1 Introduction to Systems of Linear Equations 1. Which of the following is an equation linear in 𝑥1 , 𝑥2 , 𝑥3 ? a. 2𝑥1 + 𝑥2 𝑥3 = 5 1 b. 3𝑥1 + 𝑥2 − 𝑥 = 0 3
c. √2𝑥1 − 𝑥3 = 0 d. 2𝑥1 + √𝑥3 = 1 e. None of the above Answer c. a. is not linear because you have a product of variables (𝑥2 𝑥3 ) b. is not linear because you have the reciprocal (−1 power) of the variable 𝑥3 c. is linear because √2 is just a number (another coefficient) and even thought 𝑥2 is missing, it just means that the coefficient is 0. d. is not linear because it has the square root of the variable 𝑥3 (1/2 power)
2. Which of the following is an equation linear in 𝑎, 𝑏, 𝑐? a. 2𝑎𝑥 + 3𝑏𝑦 − 4𝑐 2 𝑧 = 0 b. 3𝑎𝑥 2 + 2𝑏𝑐𝑦 = 1 2
2
2
c. 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 = 1
d. 𝑎𝑥 + 𝑏𝑦 2 + 𝑐𝑧 3 = −1 e. None of the above Answer d. *For this question, 𝑥, 𝑦, 𝑧 are not variables, they are simply numbers/constants a. not linear because of the 𝑐 2 b. not linear because of the product 𝑏𝑐 1 1 1
c. not linear because of the reciprocals (−1 powers) 𝑎 , 𝑏 , 𝑐
d. linear even though we have 𝑦 2 and 𝑧 3 because they are not variables in this question
3. Write the augmented matrix for the system of linear equations 𝑥 + 2𝑦 − 𝑧 = 0 2𝑥 − 3𝑧 = 1 2𝑦 + 5𝑧 = −1 Rewrite the system as 1𝑥 + 2𝑦 − 1𝑧 = 0 2𝑥 + 0𝑦 − 3𝑧 = 1 1𝑥 + 2𝑦 + 5𝑧 = −1 1 2 −1 0 So, the augmented matrix is [2 0 −3| 1 ] 0 2 5 −1
4. Write out the system of linear equations in the variables 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 that correspond to the following augmented matrix 0 3 0 2 0 [1 −1 1/2 0|−3] 0 0 0.6 0 0 Rewrite as equations: + 0𝑥3 + 3𝑥4 = 0 2𝑥1 + 0𝑥2 1𝑥1
−1𝑥2
0𝑥1
+0𝑥2
1
+ 2 𝑥3 + 0.6𝑥3
+0 𝑥4 + 0𝑥4
= −3 =0
2𝑥1 + 3𝑥4 = 0 1
Simplifying, we get 𝑥1 − 𝑥2 + 2 𝑥3 = −3. 0.6𝑥3 = 0
5.2 Elementary Row Operations 1. State the elementary row operation that will turn the (1,1)-entry into a 1 3 −6 12 𝐴 = [1 0 −1] 2 1 0 There are three answers that will work for this question: Option 1: swap row one and row two (𝑅1 ↔ 𝑅2 ) Option 2: add −2 times row two to row one (𝑅1 − 2𝑅2 ) Option 3: add −1 times row three to row one (𝑅1 − 𝑅3 ) 2. State the elementary row operations that will turn the (2, 1)-entry and the (3,1)-entry into 0s 0 −1 1 1 3 [ 2 −1 1 0 |−3] −3 0 −2 1 0 Add −2 times row one to row two: 0 −1 1 1 3 [ 2 −1 1 0 |−3] −3 0 −2 1 0 1 3 0 −1 1 𝑅2 :𝑅2 −2𝑅1 [ 0 −7 → 1 2 |−5] −3 0 −2 1 0 Then, add 3 times row one to row three: 𝑅3 :𝑅3 +3𝑅1
→
1 [ 0 0
3 −7 9
0 1 −2
−1 1 2 |−5] −2 3
3. State the elementary row operations that will turn 𝐴12 and 𝐴32 into a 0s 1 1 0 −1 1 [0 1 1 1 |4] 2 0 0 0 −5 Add −1 times row two to row 1: 1 1 0 −1 1 [0 1 1 1 |4] 2 0 0 0 −5 0 −1 −2 −3 𝑅1 :𝑅1 −𝑅2 1 [0 → 1 1 1| 4 ] 2 0 0 0 −5 Then, add 5 times row two to row three: 𝑅3 :𝑅3 +5𝑅2
→
1 [0 0
0 1 0
−1 1 7
−2 −3 1| 4 ] 5 20
5.3 RREF and RREF 1. Reduce the following matrix into RREF (reduced row echelon form) 3 −6 12 [1 0 −1] 2 1 0 3 [1 2
−6 0 1
1 [0 → 0 1 𝑅 1 27 3 → [0 0 𝑅2 ↔𝑅3
12 𝑅1↔𝑅2 1 0 −1 [3 −6 12 ] −1] → 0 2 1 0 𝑅2 :𝑅2 −3𝑅1 1 0 −1 𝑅3 :𝑅3 −2𝑅1 [0 −6 15 ] → 0 1 2 0 −1 𝑅3 :𝑅3+6𝑅2 1 0 −1 [0 1 2 ] 1 2 ]→ −6 15 0 0 27 +𝑅3 0 −1 𝑅𝑅1:𝑅:𝑅1−2𝑅 1 0 0 2 2 3 [0 1 0 ] 1 2 ]→ 0 1 0 0 1
2. Reduce the following matrix into RREF (reduced row echelon form) −1 3 0 −1 1 [ 2 −5 1 0 |−3] −3 0 −2 1 5 −1 3 0 −1 1 −1𝑅1 1 −3 0 1 −1 [ 2 −5 1 0 |−3] → [ 2 −5 1 0 |−3] −3 0 −2 1 5 −3 0 −2 1 5 𝑅2 :𝑅2 −2𝑅1 𝑅3 :𝑅3 +3𝑅3
1 [0 → 0 𝑅1 :𝑅1 +3𝑅2 𝑅3 :𝑅3 +9𝑅2 1 [0 → 0 𝑅1 :𝑅1 −3𝑅3 𝑅2 :𝑅2 −𝑅3 1 [0 → 0
−3 0 1 −1 1 1 −2|−1] −9 −2 4 2 1 0 3 −5 −4 7𝑅3 1 1 1 −2|−1] → [0 0 7 −14 −7 0 0 0 1 −1 1 0 0| 0 ] 0 1 −2 −1
0 1 0
3 −5 −4 1 −2|−1] 1 −2 −1
3. Reduce the following matrix into RREF (reduced row echelon form) 3 1 0 1 3 14 [2 −2 4 0 | ] 1 0 −1 2 1 5 −1 4 1 17 3 1 0 1 3 1 0 −1 2 1 14 𝑅1 ↔𝑅3 2 −2 4 0 14 [2 −2 4 0 | ] → [ | ] 1 0 −1 2 1 3 1 0 1 3 5 −1 4 1 17 5 −1 4 1 17 𝑅2 :𝑅2 −2𝑅1 0 −1 2 1 𝑅3 :𝑅3 −3𝑅1 1 𝑅4 ;𝑅4 −5𝑅1 0 −2 6 −4|12] [ → 0 1 3 −5 0 0 −1 9 −9 12 1 0 −1 2 1 1 − 𝑅2 2 1 −3 2 |−6] → [0 0 1 3 −5 0 0 −1 9 −9 12 −1 2 1 𝑅3 :𝑅3 −𝑅2 1 0 𝑅4 :𝑅4 +𝑅2 0 1 −3 2 |−6] [ → 0 0 6 −7 6 0 0 6 −7 6 1 0 −1 2 1 1 𝑅3 6 −3 2 |−6] → [0 1 0 0 1 −7/6 1 0 0 6 −7 6 𝑅1 :𝑅1 +𝑅3 5/6 2 𝑅2 :𝑅2 +3𝑅3 1 0 0 𝑅4 :𝑅4 −6𝑅3 0 1 0 −3/2 −3 [ | ] → 0 0 1 −7/6 1 0 0 0 0 0
5.4 Solutions to an SLE 1. State all the solutions (if any) to the system of linear equations that corresponds to the following augmented matrix 1 0 0 4 [0 1 0| 0 ] 0 0 1 −1 𝑥 4 x⃗ = [𝑦] = [ 0 ] 𝑧 −1 2. Find the general solution (if any) to the system of linear equations that corresponds to the following augmented matrix 1 3 0 −5 [0 0 1| 3 ] 0 0 0 0 𝑥 −3 −5 − 3𝑡 −5 ] = [ 0 ]+𝑡[ 1 ] x⃗ = [𝑦] = [ 𝑡 𝑧 0 3 3 3. Find the general solution (if any) to the system of linear equations that corresponds to the following augmented matrix 1 3 0 −5 [0 0 1| 3 ] 0 0 0 1 𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑟𝑜𝑤.
4. State all the solutions (if any) to the line system that is represented by the augmented matrix 1 −2 0 [0 0 0 0 0
0 1 0 0
33 1|5] 00 00
𝑥1 3 + 2𝑡 − 3𝑠 3 2 −3 𝑥2 𝑡 0 1 0 ] = [ ]+𝑡[ ]+𝑠[ ] x⃗ = [𝑥 ] = [ 3 5−𝑠 5 0 −1 𝑥4 𝑠 0 0 1 5. How many solutions does the system of linear equations with this augmented matrix have? 1 0 2 2 [−1 1 1|−2] 0 1 2 0 1 0 2 2 𝑅2 +𝑅1 1 0 2 2 𝑅3 −𝑅2 1 0 2 2 [−1 1 1|−2] → [ 0 1 3|0 ] → [0 1 3 |0] 0 1 2 0 0 1 20 0 0 −1 0 We don’t have to reduce this into full RREF, we already see that there will be three leading ones in the diagonal of the coefficient matrix. So, the system has a unique solution.
6. Find all 𝑘 values such that the system with the following augmented matrix has no solutions. 1 0 2 2 [−1 1 2 |−1] 0 1 𝑘2 𝑘 1 0 2 2 𝑅2+𝑅1 1 0 2 2 [−1 1 2 |0] → [0 1 4 |2] 2 0 1 𝑘 𝑘 0 1 𝑘2 𝑘 2 2 𝑅3 −𝑅2 1 0 → [0 1 4 | 2 ] 0 0 𝑘2 − 4 𝑘 − 2 For there to be no solutions, we need the last row to be a zero row with a nonzero constant. So, we need 𝑘 2 − 4 = 0 and 𝑘 − 2 ≠ 0 (𝑘 − 2)(𝑘 + 2) = 0 and 𝑘 − 2 ≠ 0 Therefore, we need 𝑘 = −2
5.5 Gaussian Elimination 2𝑥 + 4𝑦 − 2𝑧 = 5 𝑥−𝑦 =2 1. Find the general solution to the system . 3𝑥 + 3𝑦 − 2𝑧 = 7 2 4 −2 5 𝑅1 ↔𝑅2 1 −1 0 2 [1 −1 [2 0 |2] → 4 −2|5] 3 3 −2 7 3 3 −2 7 𝑅2 :𝑅2 −2𝑅1 1 −1 0 2 𝑅3 :𝑅3 −3𝑅1 [0 → 6 −2|1] 0 6 −2 1 because rows 2 and 3 are identical, let’s add −1 times row 2 to row 3 before we get the next leading one. 1 0 2 1 −1 0 2 𝑅 1 −1 𝑅3 :𝑅3 −𝑅2 6 2 [0 → 1 −1/3|1/6] 6 −2|1] → [0 0 0 0 0 0 0 0 0 1 0 −1/3 13/6 𝑅1 :𝑅1 +𝑅2 [0 1 −1/3| 1/6 ] → 0 0 0 0 Since the variable 𝑧 has no leading one, we assign it a parameter 𝑡. Therefore, the solution is 13 1 13 1 + 𝑡 𝑥 6 3 6 3 x⃗ = [𝑦] = 1 1 = 1 + 𝑡 1 + 𝑡 𝑧 6 3 6 3 [ ] [ [ ] 𝑡 0 1]
𝑥−𝑦−𝑧 =3 . 𝑧=5 1 −1 −1 3 𝑅1 :𝑅1+𝑅2 1 −1 0 8 [ | ]→ [ | ] 0 0 15 0 0 15 Since the parameter 𝑦 has no leading one, we assign it a parameter 𝑡. 𝑥 8+𝑡 8 Therefore, the solutions is x⃗ = [𝑦] = [ 𝑡 ] = [0] + 𝑧 5 5 1 𝑡 [ 1] 0
2. Find the general solution to the system
5.6 Rank 1 1. Determine the rank of the matrix 𝐴 = [0 0 0
0 0 0 0
1 2 0.5 2 −2 8 ]. 0 0 1 0 0 0
We can reduce this matrix further into RREF but we don’t have to, we already see that row 1 and row 3 have leading ones. Since row 2 has no non-zero values below it, row two will also have a leading one. So 𝑟𝑎𝑛𝑘(𝐴) = 3. 2. If a system of 5 linear equations in 4 unknowns has a general solution with 1 parameter, what is the rank of the augmented matrix? What is the rank of the coefficient matrix? ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ The matrix looks something like this ∗ ∗ ∗ ∗||∗ . The ∗ ∗ ∗ ∗∗ [∗ ∗ ∗ ∗ ∗] max rank is 4 since there are only 4 unknowns. Because the solution has one parameter, that means that it has one row with no leading one. So, there are three rows with leading ones. Therefore, the rank of the coefficient matrix is 3. Since the system is consistent (has solutions), the augmented matrix must have the same rank as the coefficient matrix. So, the augmented matrix also has rank 3.
3. Suppose that a system of 𝑚 linear equations in 𝑛 variables can be represented by [𝐴|𝑏]. If 𝑟𝑎𝑛𝑘([𝐴|𝑏]) = 𝑟𝑎𝑛𝑘(𝐴) = 𝑛, which of the following statements best describes the system? a. The system will always have no solution b. The system will always have a unique solution Since the rank of the coefficient matrix is the same as the rank of the augmented matrix, the system is consistent, there is at least one solution. Since the matrix has rank 𝑛, that means that every variable has a leading one and the matrix is of full rank. Therefore, there is a unique solution. *In this case, 𝑚 must be greater or equal to 𝑛 because we need at least 𝑛 rows to hold the 𝑛 leading ones. c. The system will always have a general solution with 1 parameter (i.e. infinitely many solutions) d. If 𝑛 > 𝑚, the system will have infinitely many solutions. e. If 𝑛 < 𝑚, the system will have infinitely many solutions.
5.7 Homogeneous System of Linear Equations 1. True or False. A homogeneous system with 𝑚 linear equations and 𝑚 variables will always have a unique solution. False. Here’s a 3 × 3 matrix in RREF that has more variables than 1 ∗ ∗0 leading ones [0 1 ∗ |0], so the solution will have one 0 0 00 parameter, meaning it will have infinitely many solutions. 2 4 −2 2. If 𝐴 = [1 −1 0 ], solve the associated homogeneous 3 3 −2 system of linear equations. 2 4 −2 0 𝑅1 ↔𝑅2 1 −1 0 0 [1 −1 0 |0] → [2 4 −2|0] 3 3 −2 0 3 3 −2 0 𝑅2 :𝑅2 −2𝑅1 1 −1 0 0 𝑅3:𝑅3−𝑅2 1 −1 0 0 𝑅3 :𝑅3 −3𝑅1 [0 6 −2|0] → [0 6 −2|0] → 0 6 −2 0 0 0 0 0 1 0 0 𝑅 :𝑅 +𝑅 1 0 −1/3 0 𝑅 1 −1 1 1 2 6 2 [0 1 −1/3|0] → [0 1 −1/3|0] → 0 0 0 0 0 0 0 0 𝑥 1/3 1/3𝑡 So, the solution is x⃗ = [𝑦] = [1/3𝑡] = 𝑡 [1/3] 𝑧 𝑡 1
End of Chapter 5 Exam-Like Quiz 1. Which of the following is in RREF (row-reduced echelon form)? 1 0 0 0 1 −1 0 0 𝐴 = [0 0 1 0] 𝐵 = [0 0 1 −1] 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 −1 0 0 1 1 0 0 1 0 ] 𝐷 = [ 0 1 1 0] 𝐶 = [0 0 0 0 0 −1 0 0 0 1 0 0 0 0 0 0 0 0 a. A and C b. B and C c. C and D d. B only e. C only Answer d. The matrix A is not in RREF because the leading ones don’t follow the “ladder formation”. The matrix B is in RREF because all zero rows are at the bottom, the first non-zero entry of each row is a leading one, the leading ones follow a downward “ladder formation”, and we have all zeros above and below each leading one. The matrix C is not in RREF because of the −1 in row three. The matrix D is not in RREF because there is a 1 above the leading one in row two.
2.
Which of the following equations is/are linear in the variables 𝑥, 𝑦, and 𝑧? i.
𝑥 − 𝑦 + 2𝑧 = 0 1
1
1
iii.
+𝑦−𝑧 =2 𝑥
v.
2𝑥 − 𝑦 = √5 − 𝑧
ii.
𝑥 2 + 2𝑦 2 + 3𝑧 2 = 1
iv.
𝑥𝑦𝑧 = 1
a. i, ii, iii and iv b. iii and v c. i only d. i and v e. i, iii, iv, and v Answer d. i. ii.
linear not linear since 𝑥, 𝑦, 𝑧 are all squared
iii.
not linear since we have reciprocal functions 𝑥 , 𝑦 , 𝑧
iv. v.
not linear since we have a product of variables 𝑥𝑦𝑧 linear. We can rearrange into standard form: 2𝑥 − 𝑦 + 𝑧 = √5 *Even though we have a √ , it’s not the square root of a variable, just a constant.
1 1 1
The following augmented matrix corresponds to a system of linear equations in the variables 𝑥, 𝑦, and 𝑧. Use it to answer questions 3-5. 1 [0 0
0 1 0
−3 −5 7 | 0 ] (𝑘 + 2)(𝑘 − 4) (𝑘 − 4)
3. For what value(s) of 𝑘 does the system have no solution? a. 𝑘 = −2 only b. 𝑘 = 4 only c. Any value such that 𝑘 ≠ −2 and 𝑘 ≠ 4 d. 𝑘 = −2 or 𝑘 = −4 e. For no values of 𝑘 Answer a. For there to be no solution, we need the last row to be a zero row with a non-zero constant. If 𝑘 = −2, the last row becomes [0 0 0 | − 6]
4. For what value(s) of 𝑘 does the system have a unique solution? a. 𝑘 = −2 only b. 𝑘 = 4 only c. Any value such that 𝑘 ≠ −2 and 𝑘 ≠ 4 d. 𝑘 = −2 or 𝑘 = −4 e. For no values of 𝑘 Answer c. For there to be a unique solution, we need the identity matrix on the left (coefficient matrix). So, (𝑘 + 2)(𝑘 − 4) ≠ 0 −𝑘 ≠ 2 or 𝑘 ≠ 4
5. For what value(s) of 𝑘 does the system have a oneparameter family of solutions (infinitely many solutions) a. 𝑘 = −2 only b. 𝑘 = 4 only c. Any value such that 𝑘 ≠ −2 and 𝑘 ≠ 4 d. 𝑘 = −2 or 𝑘 = −4 e. For no values of 𝑘 Answer b. For there to be infinitely many solutions, we need a zero row at the bottom. So, we need (𝑘 + 2)(𝑘 − 4) = 0 and 𝑘 − 4 = 0. i.e. 𝑘 = 4
6. 𝐵 is the first row of the row-reduced echelon form of the matrix 1 2 3 𝐴 = [−1 −1 0] 0 1 3 a. [1 0 0] b. [1 0 3] c. [1 0 −3] d. [1 0 −1] e. None of the above Answer c. 1 [−1 0
2 −1 1
3 𝑅2+𝑅1 1 [0 0] → 3 0
First row is [1 0 − 3]
2 1 2 3 𝑅𝑅1−2𝑅 −𝑅 3 2 [0 1 3] → 1 3 0
0 −3 1 3] 0 0
7. Suppose that a system of 𝑚 linear equations in 𝑛 variables is represented by the augmented matrix [𝐴|𝑏]. If the system is consistent, which of the following statements is always true? a. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) ≥ 𝑟𝑎𝑛𝑘(𝐴) b. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) ≤ 𝑟𝑎𝑛𝑘(𝐴) c. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) = 𝑟𝑎𝑛𝑘(𝐴) d. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) = 𝑟𝑎𝑛𝑘(𝐴) = 𝑛 e. None of the above. Answer b. For the system to be consistent, there must be a solution. Hence, we cannot end up with a zero row in the coefficients matrix and a non-zero constant. So, we don’t want the case where 𝑟𝑎𝑛𝑘(𝐴) < 𝑟𝑎𝑛𝑘([𝐴|𝑏]). Therefore, 𝑟𝑎𝑛𝑘(𝐴) ≥ 𝑟𝑎𝑛𝑘([𝐴|𝑏]). Flipping the inequality around, 𝑟𝑎𝑛𝑘([𝐴|𝑏]) ≤ 𝑟𝑎𝑛𝑘(𝐴).
8. Write the general solution (all solutions) to the system of linear equations 𝑥 − 𝑦 − 3𝑧 = −4 2𝑥 − 3𝑦 + 𝑧 = 0 a. (−12, −8, 0) b. (−12, −8, 0) + 𝑡(10, 7, 1) c. (10, 7, 1) + 𝑡(−12, −8, 0) d. (−4, 0, 0) + 𝑡(10, 7, 1) e. (−4, 0, 0) + 𝑡(−12, −8, 0) Answer b. Solving the system, we get [
1 2
1𝑅2
→ [
−1 −3 1 0
−3 −4 𝑅2:𝑅2−2𝑅1 1 | ]→ [ 1 0 0 −1 1
−1 −1
−3 −4 𝑅1:𝑅1+𝑅2 1 | ]→ [ −7 −8 0
0 1
−3 −4 | ] 7 8 −10 −12 | ] −7 −8
Since the variable 𝑧 doesn’t have a leading one, assign it the parameter 𝑡 𝑥 −12 + 10𝑡 −12 10 The solution is x⃗ = [𝑦] = [ −8 + 7𝑡 ] = [ −8 ] + 𝑡 [ 7 ] 𝑧 𝑡 0 1 Or (−12, −8, 0) + 𝑡(10, 7, 1)
9. Given a system of 8 linear equations in 5 variables, if the rank of the coefficient matrix 𝐴 is 𝑟, what is the maximum value 𝑟 can be? a. 1 b. 3 c. 5 d. 8 e. Not enough information given Answer c. The rank of a matrix is the number of leading ones in the RREF (reduced row echelon form). A matrix can only hold as many leading ones as the number of rows or the number of columns, whichever is smaller. So, in this case, since there are only 5 variables but 8 equations, the max rank can only be 5.
10. Given a system of 3 linear equations in 5 variables, if the rank of the coefficient matrix 𝐴 is 3, which of the following statements best describes the system? a. The system has no solutions b. The system has a unique solution c. The system has a general solution with 1 parameter d. The system has a general solution with 2 parameters e. None of the above Answer d. An example of the system in discussion is 1 ∗ ∗ ∗ ∗∗ [0 1 ∗ ∗ ∗|∗]. Notice that two of the 5 variables do 0 0 1 ∗ ∗∗ not have leading ones, they must be assigned parameters. Therefore, the general solution will have 2 parameters.
11. A homogeneous system (𝐴𝑥 = ⃗0) with 𝑚 linear equations in 𝑛 unknowns has exactly one solution. Which of the following statements is always true? a. The system has at least one row of zeros b. 𝑚 > 𝑛 c. 𝑚 < 𝑛 d. The RREF of 𝐴 is the identity matrix e. None of the above Answer e. a. Indeed, the system may have a row of zeros, for 1 00 example [0 1|0] (𝑚 = 3, 𝑛 = 2); however, this is not 0 00 1 00 | ] (𝑚 = 2 = 𝑛) always the case, for example [ 0 10 b. Indeed, sometimes we can have 𝑚 > 𝑛, but this is not always the case (see the examples used in a.) c. This is never true. Suppose that we did have 𝑚 < 𝑛, then the max rank will be 𝑚. So, there will be (𝑛 − 𝑚) variables with no leading ones, which means there will be parameters in our solution. (i.e. not exactly one solution) d. Indeed, the RREF can be the identity, but it doesn’t have to be. (see the example used in a.)
c. √2𝑥1 − 𝑥3 = 0 d. 2𝑥1 + √𝑥3 = 1 e. None of the above Answer c. a. is not linear because you have a product of variables (𝑥2 𝑥3 ) b. is not linear because you have the reciprocal (−1 power) of the variable 𝑥3 c. is linear because √2 is just a number (another coefficient) and even thought 𝑥2 is missing, it just means that the coefficient is 0. d. is not linear because it has the square root of the variable 𝑥3 (1/2 power)
2. Which of the following is an equation linear in 𝑎, 𝑏, 𝑐? a. 2𝑎𝑥 + 3𝑏𝑦 − 4𝑐 2 𝑧 = 0 b. 3𝑎𝑥 2 + 2𝑏𝑐𝑦 = 1 2
2
2
c. 𝑎 𝑥 + 𝑏 𝑦 + 𝑐 𝑧 = 1
d. 𝑎𝑥 + 𝑏𝑦 2 + 𝑐𝑧 3 = −1 e. None of the above Answer d. *For this question, 𝑥, 𝑦, 𝑧 are not variables, they are simply numbers/constants a. not linear because of the 𝑐 2 b. not linear because of the product 𝑏𝑐 1 1 1
c. not linear because of the reciprocals (−1 powers) 𝑎 , 𝑏 , 𝑐
d. linear even though we have 𝑦 2 and 𝑧 3 because they are not variables in this question
3. Write the augmented matrix for the system of linear equations 𝑥 + 2𝑦 − 𝑧 = 0 2𝑥 − 3𝑧 = 1 2𝑦 + 5𝑧 = −1 Rewrite the system as 1𝑥 + 2𝑦 − 1𝑧 = 0 2𝑥 + 0𝑦 − 3𝑧 = 1 1𝑥 + 2𝑦 + 5𝑧 = −1 1 2 −1 0 So, the augmented matrix is [2 0 −3| 1 ] 0 2 5 −1
4. Write out the system of linear equations in the variables 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 that correspond to the following augmented matrix 0 3 0 2 0 [1 −1 1/2 0|−3] 0 0 0.6 0 0 Rewrite as equations: + 0𝑥3 + 3𝑥4 = 0 2𝑥1 + 0𝑥2 1𝑥1
−1𝑥2
0𝑥1
+0𝑥2
1
+ 2 𝑥3 + 0.6𝑥3
+0 𝑥4 + 0𝑥4
= −3 =0
2𝑥1 + 3𝑥4 = 0 1
Simplifying, we get 𝑥1 − 𝑥2 + 2 𝑥3 = −3. 0.6𝑥3 = 0
5.2 Elementary Row Operations 1. State the elementary row operation that will turn the (1,1)-entry into a 1 3 −6 12 𝐴 = [1 0 −1] 2 1 0 There are three answers that will work for this question: Option 1: swap row one and row two (𝑅1 ↔ 𝑅2 ) Option 2: add −2 times row two to row one (𝑅1 − 2𝑅2 ) Option 3: add −1 times row three to row one (𝑅1 − 𝑅3 ) 2. State the elementary row operations that will turn the (2, 1)-entry and the (3,1)-entry into 0s 0 −1 1 1 3 [ 2 −1 1 0 |−3] −3 0 −2 1 0 Add −2 times row one to row two: 0 −1 1 1 3 [ 2 −1 1 0 |−3] −3 0 −2 1 0 1 3 0 −1 1 𝑅2 :𝑅2 −2𝑅1 [ 0 −7 → 1 2 |−5] −3 0 −2 1 0 Then, add 3 times row one to row three: 𝑅3 :𝑅3 +3𝑅1
→
1 [ 0 0
3 −7 9
0 1 −2
−1 1 2 |−5] −2 3
3. State the elementary row operations that will turn 𝐴12 and 𝐴32 into a 0s 1 1 0 −1 1 [0 1 1 1 |4] 2 0 0 0 −5 Add −1 times row two to row 1: 1 1 0 −1 1 [0 1 1 1 |4] 2 0 0 0 −5 0 −1 −2 −3 𝑅1 :𝑅1 −𝑅2 1 [0 → 1 1 1| 4 ] 2 0 0 0 −5 Then, add 5 times row two to row three: 𝑅3 :𝑅3 +5𝑅2
→
1 [0 0
0 1 0
−1 1 7
−2 −3 1| 4 ] 5 20
5.3 RREF and RREF 1. Reduce the following matrix into RREF (reduced row echelon form) 3 −6 12 [1 0 −1] 2 1 0 3 [1 2
−6 0 1
1 [0 → 0 1 𝑅 1 27 3 → [0 0 𝑅2 ↔𝑅3
12 𝑅1↔𝑅2 1 0 −1 [3 −6 12 ] −1] → 0 2 1 0 𝑅2 :𝑅2 −3𝑅1 1 0 −1 𝑅3 :𝑅3 −2𝑅1 [0 −6 15 ] → 0 1 2 0 −1 𝑅3 :𝑅3+6𝑅2 1 0 −1 [0 1 2 ] 1 2 ]→ −6 15 0 0 27 +𝑅3 0 −1 𝑅𝑅1:𝑅:𝑅1−2𝑅 1 0 0 2 2 3 [0 1 0 ] 1 2 ]→ 0 1 0 0 1
2. Reduce the following matrix into RREF (reduced row echelon form) −1 3 0 −1 1 [ 2 −5 1 0 |−3] −3 0 −2 1 5 −1 3 0 −1 1 −1𝑅1 1 −3 0 1 −1 [ 2 −5 1 0 |−3] → [ 2 −5 1 0 |−3] −3 0 −2 1 5 −3 0 −2 1 5 𝑅2 :𝑅2 −2𝑅1 𝑅3 :𝑅3 +3𝑅3
1 [0 → 0 𝑅1 :𝑅1 +3𝑅2 𝑅3 :𝑅3 +9𝑅2 1 [0 → 0 𝑅1 :𝑅1 −3𝑅3 𝑅2 :𝑅2 −𝑅3 1 [0 → 0
−3 0 1 −1 1 1 −2|−1] −9 −2 4 2 1 0 3 −5 −4 7𝑅3 1 1 1 −2|−1] → [0 0 7 −14 −7 0 0 0 1 −1 1 0 0| 0 ] 0 1 −2 −1
0 1 0
3 −5 −4 1 −2|−1] 1 −2 −1
3. Reduce the following matrix into RREF (reduced row echelon form) 3 1 0 1 3 14 [2 −2 4 0 | ] 1 0 −1 2 1 5 −1 4 1 17 3 1 0 1 3 1 0 −1 2 1 14 𝑅1 ↔𝑅3 2 −2 4 0 14 [2 −2 4 0 | ] → [ | ] 1 0 −1 2 1 3 1 0 1 3 5 −1 4 1 17 5 −1 4 1 17 𝑅2 :𝑅2 −2𝑅1 0 −1 2 1 𝑅3 :𝑅3 −3𝑅1 1 𝑅4 ;𝑅4 −5𝑅1 0 −2 6 −4|12] [ → 0 1 3 −5 0 0 −1 9 −9 12 1 0 −1 2 1 1 − 𝑅2 2 1 −3 2 |−6] → [0 0 1 3 −5 0 0 −1 9 −9 12 −1 2 1 𝑅3 :𝑅3 −𝑅2 1 0 𝑅4 :𝑅4 +𝑅2 0 1 −3 2 |−6] [ → 0 0 6 −7 6 0 0 6 −7 6 1 0 −1 2 1 1 𝑅3 6 −3 2 |−6] → [0 1 0 0 1 −7/6 1 0 0 6 −7 6 𝑅1 :𝑅1 +𝑅3 5/6 2 𝑅2 :𝑅2 +3𝑅3 1 0 0 𝑅4 :𝑅4 −6𝑅3 0 1 0 −3/2 −3 [ | ] → 0 0 1 −7/6 1 0 0 0 0 0
5.4 Solutions to an SLE 1. State all the solutions (if any) to the system of linear equations that corresponds to the following augmented matrix 1 0 0 4 [0 1 0| 0 ] 0 0 1 −1 𝑥 4 x⃗ = [𝑦] = [ 0 ] 𝑧 −1 2. Find the general solution (if any) to the system of linear equations that corresponds to the following augmented matrix 1 3 0 −5 [0 0 1| 3 ] 0 0 0 0 𝑥 −3 −5 − 3𝑡 −5 ] = [ 0 ]+𝑡[ 1 ] x⃗ = [𝑦] = [ 𝑡 𝑧 0 3 3 3. Find the general solution (if any) to the system of linear equations that corresponds to the following augmented matrix 1 3 0 −5 [0 0 1| 3 ] 0 0 0 1 𝑁𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑟𝑜𝑤.
4. State all the solutions (if any) to the line system that is represented by the augmented matrix 1 −2 0 [0 0 0 0 0
0 1 0 0
33 1|5] 00 00
𝑥1 3 + 2𝑡 − 3𝑠 3 2 −3 𝑥2 𝑡 0 1 0 ] = [ ]+𝑡[ ]+𝑠[ ] x⃗ = [𝑥 ] = [ 3 5−𝑠 5 0 −1 𝑥4 𝑠 0 0 1 5. How many solutions does the system of linear equations with this augmented matrix have? 1 0 2 2 [−1 1 1|−2] 0 1 2 0 1 0 2 2 𝑅2 +𝑅1 1 0 2 2 𝑅3 −𝑅2 1 0 2 2 [−1 1 1|−2] → [ 0 1 3|0 ] → [0 1 3 |0] 0 1 2 0 0 1 20 0 0 −1 0 We don’t have to reduce this into full RREF, we already see that there will be three leading ones in the diagonal of the coefficient matrix. So, the system has a unique solution.
6. Find all 𝑘 values such that the system with the following augmented matrix has no solutions. 1 0 2 2 [−1 1 2 |−1] 0 1 𝑘2 𝑘 1 0 2 2 𝑅2+𝑅1 1 0 2 2 [−1 1 2 |0] → [0 1 4 |2] 2 0 1 𝑘 𝑘 0 1 𝑘2 𝑘 2 2 𝑅3 −𝑅2 1 0 → [0 1 4 | 2 ] 0 0 𝑘2 − 4 𝑘 − 2 For there to be no solutions, we need the last row to be a zero row with a nonzero constant. So, we need 𝑘 2 − 4 = 0 and 𝑘 − 2 ≠ 0 (𝑘 − 2)(𝑘 + 2) = 0 and 𝑘 − 2 ≠ 0 Therefore, we need 𝑘 = −2
5.5 Gaussian Elimination 2𝑥 + 4𝑦 − 2𝑧 = 5 𝑥−𝑦 =2 1. Find the general solution to the system . 3𝑥 + 3𝑦 − 2𝑧 = 7 2 4 −2 5 𝑅1 ↔𝑅2 1 −1 0 2 [1 −1 [2 0 |2] → 4 −2|5] 3 3 −2 7 3 3 −2 7 𝑅2 :𝑅2 −2𝑅1 1 −1 0 2 𝑅3 :𝑅3 −3𝑅1 [0 → 6 −2|1] 0 6 −2 1 because rows 2 and 3 are identical, let’s add −1 times row 2 to row 3 before we get the next leading one. 1 0 2 1 −1 0 2 𝑅 1 −1 𝑅3 :𝑅3 −𝑅2 6 2 [0 → 1 −1/3|1/6] 6 −2|1] → [0 0 0 0 0 0 0 0 0 1 0 −1/3 13/6 𝑅1 :𝑅1 +𝑅2 [0 1 −1/3| 1/6 ] → 0 0 0 0 Since the variable 𝑧 has no leading one, we assign it a parameter 𝑡. Therefore, the solution is 13 1 13 1 + 𝑡 𝑥 6 3 6 3 x⃗ = [𝑦] = 1 1 = 1 + 𝑡 1 + 𝑡 𝑧 6 3 6 3 [ ] [ [ ] 𝑡 0 1]
𝑥−𝑦−𝑧 =3 . 𝑧=5 1 −1 −1 3 𝑅1 :𝑅1+𝑅2 1 −1 0 8 [ | ]→ [ | ] 0 0 15 0 0 15 Since the parameter 𝑦 has no leading one, we assign it a parameter 𝑡. 𝑥 8+𝑡 8 Therefore, the solutions is x⃗ = [𝑦] = [ 𝑡 ] = [0] + 𝑧 5 5 1 𝑡 [ 1] 0
2. Find the general solution to the system
5.6 Rank 1 1. Determine the rank of the matrix 𝐴 = [0 0 0
0 0 0 0
1 2 0.5 2 −2 8 ]. 0 0 1 0 0 0
We can reduce this matrix further into RREF but we don’t have to, we already see that row 1 and row 3 have leading ones. Since row 2 has no non-zero values below it, row two will also have a leading one. So 𝑟𝑎𝑛𝑘(𝐴) = 3. 2. If a system of 5 linear equations in 4 unknowns has a general solution with 1 parameter, what is the rank of the augmented matrix? What is the rank of the coefficient matrix? ∗ ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ The matrix looks something like this ∗ ∗ ∗ ∗||∗ . The ∗ ∗ ∗ ∗∗ [∗ ∗ ∗ ∗ ∗] max rank is 4 since there are only 4 unknowns. Because the solution has one parameter, that means that it has one row with no leading one. So, there are three rows with leading ones. Therefore, the rank of the coefficient matrix is 3. Since the system is consistent (has solutions), the augmented matrix must have the same rank as the coefficient matrix. So, the augmented matrix also has rank 3.
3. Suppose that a system of 𝑚 linear equations in 𝑛 variables can be represented by [𝐴|𝑏]. If 𝑟𝑎𝑛𝑘([𝐴|𝑏]) = 𝑟𝑎𝑛𝑘(𝐴) = 𝑛, which of the following statements best describes the system? a. The system will always have no solution b. The system will always have a unique solution Since the rank of the coefficient matrix is the same as the rank of the augmented matrix, the system is consistent, there is at least one solution. Since the matrix has rank 𝑛, that means that every variable has a leading one and the matrix is of full rank. Therefore, there is a unique solution. *In this case, 𝑚 must be greater or equal to 𝑛 because we need at least 𝑛 rows to hold the 𝑛 leading ones. c. The system will always have a general solution with 1 parameter (i.e. infinitely many solutions) d. If 𝑛 > 𝑚, the system will have infinitely many solutions. e. If 𝑛 < 𝑚, the system will have infinitely many solutions.
5.7 Homogeneous System of Linear Equations 1. True or False. A homogeneous system with 𝑚 linear equations and 𝑚 variables will always have a unique solution. False. Here’s a 3 × 3 matrix in RREF that has more variables than 1 ∗ ∗0 leading ones [0 1 ∗ |0], so the solution will have one 0 0 00 parameter, meaning it will have infinitely many solutions. 2 4 −2 2. If 𝐴 = [1 −1 0 ], solve the associated homogeneous 3 3 −2 system of linear equations. 2 4 −2 0 𝑅1 ↔𝑅2 1 −1 0 0 [1 −1 0 |0] → [2 4 −2|0] 3 3 −2 0 3 3 −2 0 𝑅2 :𝑅2 −2𝑅1 1 −1 0 0 𝑅3:𝑅3−𝑅2 1 −1 0 0 𝑅3 :𝑅3 −3𝑅1 [0 6 −2|0] → [0 6 −2|0] → 0 6 −2 0 0 0 0 0 1 0 0 𝑅 :𝑅 +𝑅 1 0 −1/3 0 𝑅 1 −1 1 1 2 6 2 [0 1 −1/3|0] → [0 1 −1/3|0] → 0 0 0 0 0 0 0 0 𝑥 1/3 1/3𝑡 So, the solution is x⃗ = [𝑦] = [1/3𝑡] = 𝑡 [1/3] 𝑧 𝑡 1
End of Chapter 5 Exam-Like Quiz 1. Which of the following is in RREF (row-reduced echelon form)? 1 0 0 0 1 −1 0 0 𝐴 = [0 0 1 0] 𝐵 = [0 0 1 −1] 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 −1 0 0 1 1 0 0 1 0 ] 𝐷 = [ 0 1 1 0] 𝐶 = [0 0 0 0 0 −1 0 0 0 1 0 0 0 0 0 0 0 0 a. A and C b. B and C c. C and D d. B only e. C only Answer d. The matrix A is not in RREF because the leading ones don’t follow the “ladder formation”. The matrix B is in RREF because all zero rows are at the bottom, the first non-zero entry of each row is a leading one, the leading ones follow a downward “ladder formation”, and we have all zeros above and below each leading one. The matrix C is not in RREF because of the −1 in row three. The matrix D is not in RREF because there is a 1 above the leading one in row two.
2.
Which of the following equations is/are linear in the variables 𝑥, 𝑦, and 𝑧? i.
𝑥 − 𝑦 + 2𝑧 = 0 1
1
1
iii.
+𝑦−𝑧 =2 𝑥
v.
2𝑥 − 𝑦 = √5 − 𝑧
ii.
𝑥 2 + 2𝑦 2 + 3𝑧 2 = 1
iv.
𝑥𝑦𝑧 = 1
a. i, ii, iii and iv b. iii and v c. i only d. i and v e. i, iii, iv, and v Answer d. i. ii.
linear not linear since 𝑥, 𝑦, 𝑧 are all squared
iii.
not linear since we have reciprocal functions 𝑥 , 𝑦 , 𝑧
iv. v.
not linear since we have a product of variables 𝑥𝑦𝑧 linear. We can rearrange into standard form: 2𝑥 − 𝑦 + 𝑧 = √5 *Even though we have a √ , it’s not the square root of a variable, just a constant.
1 1 1
The following augmented matrix corresponds to a system of linear equations in the variables 𝑥, 𝑦, and 𝑧. Use it to answer questions 3-5. 1 [0 0
0 1 0
−3 −5 7 | 0 ] (𝑘 + 2)(𝑘 − 4) (𝑘 − 4)
3. For what value(s) of 𝑘 does the system have no solution? a. 𝑘 = −2 only b. 𝑘 = 4 only c. Any value such that 𝑘 ≠ −2 and 𝑘 ≠ 4 d. 𝑘 = −2 or 𝑘 = −4 e. For no values of 𝑘 Answer a. For there to be no solution, we need the last row to be a zero row with a non-zero constant. If 𝑘 = −2, the last row becomes [0 0 0 | − 6]
4. For what value(s) of 𝑘 does the system have a unique solution? a. 𝑘 = −2 only b. 𝑘 = 4 only c. Any value such that 𝑘 ≠ −2 and 𝑘 ≠ 4 d. 𝑘 = −2 or 𝑘 = −4 e. For no values of 𝑘 Answer c. For there to be a unique solution, we need the identity matrix on the left (coefficient matrix). So, (𝑘 + 2)(𝑘 − 4) ≠ 0 −𝑘 ≠ 2 or 𝑘 ≠ 4
5. For what value(s) of 𝑘 does the system have a oneparameter family of solutions (infinitely many solutions) a. 𝑘 = −2 only b. 𝑘 = 4 only c. Any value such that 𝑘 ≠ −2 and 𝑘 ≠ 4 d. 𝑘 = −2 or 𝑘 = −4 e. For no values of 𝑘 Answer b. For there to be infinitely many solutions, we need a zero row at the bottom. So, we need (𝑘 + 2)(𝑘 − 4) = 0 and 𝑘 − 4 = 0. i.e. 𝑘 = 4
6. 𝐵 is the first row of the row-reduced echelon form of the matrix 1 2 3 𝐴 = [−1 −1 0] 0 1 3 a. [1 0 0] b. [1 0 3] c. [1 0 −3] d. [1 0 −1] e. None of the above Answer c. 1 [−1 0
2 −1 1
3 𝑅2+𝑅1 1 [0 0] → 3 0
First row is [1 0 − 3]
2 1 2 3 𝑅𝑅1−2𝑅 −𝑅 3 2 [0 1 3] → 1 3 0
0 −3 1 3] 0 0
7. Suppose that a system of 𝑚 linear equations in 𝑛 variables is represented by the augmented matrix [𝐴|𝑏]. If the system is consistent, which of the following statements is always true? a. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) ≥ 𝑟𝑎𝑛𝑘(𝐴) b. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) ≤ 𝑟𝑎𝑛𝑘(𝐴) c. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) = 𝑟𝑎𝑛𝑘(𝐴) d. 𝑟𝑎𝑛𝑘([𝐴|𝑏]) = 𝑟𝑎𝑛𝑘(𝐴) = 𝑛 e. None of the above. Answer b. For the system to be consistent, there must be a solution. Hence, we cannot end up with a zero row in the coefficients matrix and a non-zero constant. So, we don’t want the case where 𝑟𝑎𝑛𝑘(𝐴) < 𝑟𝑎𝑛𝑘([𝐴|𝑏]). Therefore, 𝑟𝑎𝑛𝑘(𝐴) ≥ 𝑟𝑎𝑛𝑘([𝐴|𝑏]). Flipping the inequality around, 𝑟𝑎𝑛𝑘([𝐴|𝑏]) ≤ 𝑟𝑎𝑛𝑘(𝐴).
8. Write the general solution (all solutions) to the system of linear equations 𝑥 − 𝑦 − 3𝑧 = −4 2𝑥 − 3𝑦 + 𝑧 = 0 a. (−12, −8, 0) b. (−12, −8, 0) + 𝑡(10, 7, 1) c. (10, 7, 1) + 𝑡(−12, −8, 0) d. (−4, 0, 0) + 𝑡(10, 7, 1) e. (−4, 0, 0) + 𝑡(−12, −8, 0) Answer b. Solving the system, we get [
1 2
1𝑅2
→ [
−1 −3 1 0
−3 −4 𝑅2:𝑅2−2𝑅1 1 | ]→ [ 1 0 0 −1 1
−1 −1
−3 −4 𝑅1:𝑅1+𝑅2 1 | ]→ [ −7 −8 0
0 1
−3 −4 | ] 7 8 −10 −12 | ] −7 −8
Since the variable 𝑧 doesn’t have a leading one, assign it the parameter 𝑡 𝑥 −12 + 10𝑡 −12 10 The solution is x⃗ = [𝑦] = [ −8 + 7𝑡 ] = [ −8 ] + 𝑡 [ 7 ] 𝑧 𝑡 0 1 Or (−12, −8, 0) + 𝑡(10, 7, 1)
9. Given a system of 8 linear equations in 5 variables, if the rank of the coefficient matrix 𝐴 is 𝑟, what is the maximum value 𝑟 can be? a. 1 b. 3 c. 5 d. 8 e. Not enough information given Answer c. The rank of a matrix is the number of leading ones in the RREF (reduced row echelon form). A matrix can only hold as many leading ones as the number of rows or the number of columns, whichever is smaller. So, in this case, since there are only 5 variables but 8 equations, the max rank can only be 5.
10. Given a system of 3 linear equations in 5 variables, if the rank of the coefficient matrix 𝐴 is 3, which of the following statements best describes the system? a. The system has no solutions b. The system has a unique solution c. The system has a general solution with 1 parameter d. The system has a general solution with 2 parameters e. None of the above Answer d. An example of the system in discussion is 1 ∗ ∗ ∗ ∗∗ [0 1 ∗ ∗ ∗|∗]. Notice that two of the 5 variables do 0 0 1 ∗ ∗∗ not have leading ones, they must be assigned parameters. Therefore, the general solution will have 2 parameters.
11. A homogeneous system (𝐴𝑥 = ⃗0) with 𝑚 linear equations in 𝑛 unknowns has exactly one solution. Which of the following statements is always true? a. The system has at least one row of zeros b. 𝑚 > 𝑛 c. 𝑚 < 𝑛 d. The RREF of 𝐴 is the identity matrix e. None of the above Answer e. a. Indeed, the system may have a row of zeros, for 1 00 example [0 1|0] (𝑚 = 3, 𝑛 = 2); however, this is not 0 00 1 00 | ] (𝑚 = 2 = 𝑛) always the case, for example [ 0 10 b. Indeed, sometimes we can have 𝑚 > 𝑛, but this is not always the case (see the examples used in a.) c. This is never true. Suppose that we did have 𝑚 < 𝑛, then the max rank will be 𝑚. So, there will be (𝑛 − 𝑚) variables with no leading ones, which means there will be parameters in our solution. (i.e. not exactly one solution) d. Indeed, the RREF can be the identity, but it doesn’t have to be. (see the example used in a.)
