Chapter 11-5: Determinant and Its

Properties

Chapter Guidelines: • Determinant is a value that indicates many properties of a matrix. • Because so, determinants have a lot of applications to use.

Before we start, I hope we can discuss about the element notations we have been using so far, known as 푎푚푛, which we also used when expressing cofactors and minors. The inconvenience in this notation is that once 푚 or 푛 exceeds 10, the recording of notation becomes difficult to comprehend. 푡ℎ 푡ℎ For example, the element of a matrix at 10 row and 11 column would be expressed as 푎1011 according to our current denotation rules, but it will be hard to recognize whether 푎1011 actually means the element of at 10푡ℎ row and 11푡ℎ column or at 101푡ℎ row and 1푠푡 column. To solve that problem, mathematicians often use the notation 푎푚,푛 rather than 푎푚푛, where the column between 푚 and 푛 would help distinguish the values of 푚 and 푛.

Therefore, I will start using the notation 푎푚,푛 from now on, since it is a clearer notation to use. 1: Definition and Calculation of Determinant

Determinant is a quantity that is calculated by adding and subtracting products of specific elements in square matrices, and relates to several properties of a matrix, and qualities of applications of matrices we will see in Precalculus. For example, the determinant of a non-invertible matrix is always 0, while the determinant of an invertible matrix is never zero. Determinants can also be used to solve systems of equations.

In common usage, the determinant of a matrix 퐴 is either det⁡(퐴) or |퐴|. Meanwhile, since mathematicians can only calculate the determinant of square matrices, we will introduce the calculation of determinants based on the size of square matrices, first introducing the determinant of 2 × 2 square matrix:

푎 푏 The determinant of a 2 × 2 square matrix 퐴 = [ ] can be computed as 푎푑 − 푏푐 . 푐 푑

Based on the above definition, we can look into the determinant of a 3 × 3 square matrix:

The determinant of matrices larger than or equal to 3 × 3 is equivalent to the sum products of top-row elements and their cofactors.

To understand the above statements, we need to know these vocabularies first: cofactors and minors. We will discuss about minors first.

A minor is the determinant of a matrix resulting by excluding the column and row that a specific element 푎푖,푗 is included in, often denoted as 푀푖,푗.

I am to demonstrate the visualization and calculation of minors of 푎11, 푎12, and 푎13 for the 1 2 3 matrix 퐴 = [2 3 4] in below section: 3 4 1

In the below visualization, the bolded, green-texted element is the element this minor belongs to, which we will call 푎푖푗.

Matrix of 푀1,1 Matrix of 푀1,2 Matrix of 푀1,3

ퟏ 2 3 1 ퟐ 3 1 2 ퟑ 3 4 2 4 2 3 [2 3 4] = [ ] [2 3 4] = [ ] [2 3 4] = [ ] 4 1 3 1 3 4 3 4 1 3 4 1 3 4 1

Elements with green highlights are elements of same row or column as 푎푖,푗 does. Meanwhile, the right-side of equation gives the matrix of 푎푖,푗’s minor that we calculate determinant of. 3 4 The minor for element 푎 in this matrix will be det ([ ]) = 3 − 16 = −13. 1,1 4 1 2 4 Meanwhile the minor for element 푎 is det ([ ]) = 2 − 12 = −10. 1,2 3 1 2 3 The minor for element 푎1,3 is det ([ ]) = 8 − 9 = −1. 3 4

Now we will discuss cofactors. Here is a definition provided below:

A cofactor is the product of an element 푎푖,푗’s place sign and its minor, denoted as 퐶푖,푗

We will discuss the definition and formulation of “place signs”, a term we mentioned in 1 2 3 cofactor’s definition with, again, the matrix 퐴 = [2 3 4] : 3 4 1

Place signs are positive/negative signs distributed at specific positions in the matrix array that follows a similar pattern to a checkerboard: Adjacent to every place sign is the opposite of each other, just like adjacent to all black checks in checkerboard are only white checks. + − + − + − + − + − + The place sign for a 3 × 3 square matrix is: [− + −]; that for a 4 × 4 matrix is: [ ]. + − + − + − + − + − + If we observe closer, we will find the positive place signs only appear at elements 푎푖,푗 where 푖 + 푗 is even, while negative place signs appear only at elements 푎푖,푗 where 푖 + 푗 is odd. Therefore, the place sign of an element in square matrix can be summarized into the following 푖+푗 formula: 푃푙푎푐푒⁡푠푖푔푛⁡표푓⁡푎푖,푗 = (−1) .

푖+푗 The cofactor of an element 푎푖푗 is summarized by a formula 퐶푖,푗 = (−1) 푀푖,푗. 1+1 1+2 1+3 퐶1,1 = (−1) × (3 − 16) = −13, 퐶1,2 = (−1) × (2 − 12) = 10, 퐶1,3 = (−1) × (8 − 9) = −1.

Finally, having observed and understood terminologies mentioned in the definition of determinants for ≥ 3 × 3 square matrices, we are to revisit the definition below.

“The determinant of matrices larger than or equal to 3 × 3 is equivalent to the sum products of top- row elements and their cofactors, which are products of top-row elements’ place signs and their minors.”

This statement indicates the formula: det(푛 × 푛⁡푠푖푧푒푑⁡푚푎푡푟푖푥⁡퐴) = 푎1,1 × 퐶1,1 + 푎1,2 × 퐶1,2 + ⋯ + 푖+푗 푡ℎ 푎1,푛 × 퐶1,푛, while 퐶푖푗 = (−1) 푀푖,푗, while 푀푖,푗 is the determinant of matrix 퐴 excluding the 푖 row and 푗푡ℎ column.

1 2 3 So, for matrix 퐴 = [2 3 4]: 3 4 1 푖+푗 det(퐴) = 푎1,1 × 퐶1,1 + 푎1,2 × 퐶1,2 + 푎1,3 × 퐶1,3, where 퐶푖,푗 = (−1) 푀푖,푗. For 푚푖푛표푟푠 : 푀1,1 = (3 − 16) = −13, 푀1,2 = (2 − 12) = −10, 푀1,3 = (8 − 9) = −1. For 푐표푓푎푐푡표푟푠 : 1+1 1+2 1+3 퐶1,1 = (−1) × (3 − 16) = −13, 퐶1,2 = (−1) × (2 − 12) = 10, 퐶1,3 = (−1) × (8 − 9) = −1. Last but not least, for 푑푒푡푒푟푚푖푛푎푛푡 : 푎 × 퐶 + 푎 × 퐶 + 푎 × 퐶 = 1 × −13 + 2 × 10 + 3 × −1 = 4, so det(퐴) = 4. 1,1 1,1 1,2 1,2 1,3 1,3

Based on the definition of determinants for matrices, we can develop a formula for 3 × 3 square matrices’ determinants: 푎 푏 푐 For a matrix 퐴 = [푑 푒 푓], det(퐴) = 푎(푒푖 − 푓ℎ) − 푏(푑푖 − 푓푔) + 푐(푑ℎ − 푒푔). 푔 ℎ 푖

With above knowledges, we can finally develop a definition of determinant and its calculations:

Determinant is a quantity that only square matrices possess and is calculated as the sum of products of top row elements and their respective cofactors, while determinant for a 1 × 1 matrix is its element’s value.

Meanwhile, there are some significant properties of determinants:

1. det(퐼) = 1 2. If two rows were exchanged in a matrix, its determinant becomes its opposite value. 3. If one row was multiplied by a number in matrix, its determinant is multiplied by that number as well. The addition and multiplication in matrices reflect to its determinants linearly. 푎 ⁡푏 푎푥 ⁡푏푥 푎 ⁡푏 푎 ⁡푒 푎 ⁡푏 + 푒 EX: 푥 × det ([ ]) = det ([ ]), det ([ ]) + det ([ ]) = det ([ ]). 푐 푑 푐 푑 푐 푑 푐 푓 푐 푑 + 푓 4. Addition and subtractions of rows within a matrix does not affect its determinant. 5. det(퐴퐵) = det⁡(퐴) × det⁡(퐵).

2: Application of Determinants: Finding Inverse Matrix

We can also use the determinant of a matrix to find its inverse matrix by operating the following steps. In the following section, we will define each step and provide an example simultaneously 1 2 3 1 2 in description columns with the matrix 퐴 = [ ] and 퐵 = [2 3 4]. 3 4 3 4 1

Now we will demonstrate Step 1 of finding inverse matrices: Replacement by minors.

This step demands the mathematician (you, in this case) to replace each element of the matrix by its minor. 1 2 3 −13 −10 −1 1 2 4 3 Operations: [ ] → [ ], [2 3 4] → [−10 0 −2]. 3 4 2 1 3 4 1 −1 −2 −1

Step 2: Replacement by Cofactors

This step demands the mathematician (you again) to replace each element of the matrix after step 1 by its cofactor. −13 −10 −1 −13 10 −1 4 3 4 −3 Operations: [ ] → [ ], [−10 0 −2] → [ 10 0 2 ]. 2 1 −2 1 −1 −2 −1 −1 2 −1

Step 3: Adjugate Matrix

Now, imagine that the diagonal running up-left to down-right is an axis of reflection and exchange elements that are at the corresponding location across halves of the matrix divided by the aforementioned diagonal. This step produces the adjugate matrix, or adjoint of original matrix. −13 10 −1 −13 10 −1 4 −3 4 −2 Operations: [ ] → [ ], [ 10 0 2 ] → [ 10 0 2 ]. −2 1 −3 1 −1 2 −1 −1 2 −1

Step 4: Multiply the entire matrix by the reciprocal of original matrix’s determinant

This operation is what it literally says. Operations: −13 10 −1 −13 10 −1 4 −2 1 4 −2 1 [ ] → [ ], [ 10 0 2 ] → [ 10 0 2 ]. −3 1 −2 −3 1 4 −1 2 −1 −1 2 −1

And I would add a Step 5: Verification, using the specific property 퐴−1퐴 = 퐴퐴−1 = 퐼.

Operations: 1 2 1 4 −2 −2 + 3 1 − 1 1 0 For the 2 × 2 matrix: [ ] × [ ] = [ ] = [ ]. 3 4 −2 −3 1 −6 + 6 3 − 2 0 1 −13 10 −1 1 For the 3 × 3 matrix: [ 10 0 2 ]. 4 −1 2 −1

3: Application of Determinants: Cramer’s Rule

We have previously mentioned that we can use determinants to solve a system of equations, and the operation that enables this solution is called “Cramer’s Rule”.

Cramer’s Rule relates to the utilization of augmented matrix and expresses the solutions of a system of linear equations in terms of determinants.

푎1푥 + 푏1푦 + 푐1푧 = 푑1 푎1 푏1 푐1 푑1 A system {푎2푥 + 푏2푦 + 푐2푧 = 푑2 has an augmented matrix [푎2 푏2 푐2|푑2]. 푎3푥 + 푏3푦 + 푐3푧 = 푑3 푎3 푏3 푐3 푑3 Cramer Rule defined some denotations to represent a system’s solutions. For example, notation 퐷 stands for the determinant of coefficient matrix. Meanwhile, notation 퐷푚, with 푚 being a variable, is the determinant of the coefficient matrix with the column representing 푚’s coefficients replaced by the right argument of augmented matrix, which is the row of 푑s. 푑1 푏1 푐1 푎1 푑1 푐1 Therefore, 퐷푥 = det ([푑2 푏2 푐2]), and 퐷푦 = det ([푎2 푑2 푐2]). 푑3 푏3 푐3 푎3 푑3 푐3 퐷 Cramer’s Rule states that the solution for variable 푚 = 푚. Thus, the solution to the above system 퐷 퐷 퐷 퐷 would be ( 푥 , 푦 , 푍). 퐷 퐷 퐷

The derivation of Cramer’s Rule follows, but I would like to add a few notes onto the following derivation. The derivation below is mostly based on multiple Internet sources, which I later translate and summarize into the following, so some actions may be lost in translation. Meanwhile, it is extremely reasonable to get confused by it, as the following contents are preserved for those who are interested in finding out more about how the learned knowledge work. It is fine to be confused. Process it through a couple times if needed.

We will first construct a basic understanding about getting solutions of systems and establish some basic symbols:

“Cramer’s Rule relates to the utilization of augmented matrix and expresses the solutions of a system of linear equations in terms of determinants.” This means we can reuse something we learned in the last chapter: write an entire linear system as the expression 퐴푥 = 퐵, where 퐵 is the right argument of augmented matrix for the system and 퐴 is the coefficient matrix, and 푥 is the single-column matrix that represents variables of system. So say that we have 푘 variables to solve for, it would make 퐴 an 푘 × 푘 square matrix, and both 푋 and 퐵 are 푘 × 1 matrices. To figure out 푋, we have to attain the product 퐴−1퐵, and not the other arrangement where 퐵 is the multiplicand because it would not cause a multiplication between matrices 퐴−1 and 퐵.

In this derivation, let’s call our variables 푥1, 푥2, 푥3 … 푥푘 instead of calling them 푥, 푦, 푧 …, and define matrix 푋푛 as an altered version of 퐼푘 where column 푛 is replaced by a column of variables and 푛 ≤ 푘.

1 푥1 0 1 0 푥1 For example, for 푘 = 3, 푋2 will be [0 푥2 0], and 푋3 will be [0 1 푥2]. 0 푥3 1 0 0 푥3 퐷 Meanwhile, since in Cramer’s rule a variable 푚 = 푚, we will define a matrix 푀 that has the 퐷 푚 determinant 퐷 by being the altered version of 퐴 whose 푚푡ℎ column is replaced by 퐵’s only column. 푚

Now we will go on with the complicated parts:

푎11 ⋯ 푎1푘 If we multiply 2 matrices in the arrangement 퐴 × 푋푛, where 퐴 = [ ⋮ ⋱ ⋮ ] and 푋푛 is an 푎푘1 ⋯ 푎푘푘 푡ℎ altered version of 퐼푘 whose 푛 column becomes 푋, the single-column matrix of all variables, then the product 퐴푋푛 would have every column other than column 푛 be the same as its corresponding column with matrix 퐴, while its 푛푡ℎ column becomes exactly what the also single-columned matrix 퐵 is. This means 퐴푋푛 = 푀푛. The equation 퐴푋 = 푀 tells us |퐴||푋 | = |푀 |, based on the properties of determinant. 푛 푛 푛 푛

So what is the determinant of matrix 푋푛?

푡ℎ The matrix 푋푛 is an altered version of identity matrix, such that the 푛 column is the contents of 푡ℎ 푡ℎ matrix 푋 and the 푛 row consists of 0s and 푥푛 at the 푛 column. This means that the determinant of matrix 푋푛, which is calculated through sums of cofactors and elements of specific row, would only be 푟표푤+푐표푙푢푚푛 (−1) 푥푛 × Mn,n where 푛 ∈ ℤ. The sum of row number and column number is 2푛, which must be an even integer since 푛 ∈ ℤ. 푟표푤+푐표푙푢푚푛 2푛 This makes (−1) = (−1) = 1. Meanwhile, 푀푛,푛 that is in fact 푥푛 per se, would be equivalent to the determinant of identity matrix. This can be demonstrated by writing out the matrix 푋푛, then crossing out the column and row 푎푛,푛 locates at. So in conclusion, |푋 | = 푥 . 푛 푛

푡ℎ 푡ℎ To clarify on the minor of element at 푛 row and 푛 column of 푋푛, I will construct the matrix of 푋푛 below.

1 ⋯ 푥1 ⋯ 0 ⋮ ⋱ ⋮ ⋱ ⋮

푋푛 = 0 ⋯ 푥푛 ⋯ 0 ⋮ ⋱ ⋮ ⋱ ⋮ [0 ⋯ 푥푘 ⋯ 1] Once the highlighted row and column are eliminated to find minor, we can notice that finding minor is equivalent to finding the determinant of an identity matrix. Then, since det(퐼) = 1, 푀푛,푛 = 1.

So in conclusion:

퐴푋푛 = 푀푛, which means det⁡(퐴) × det⁡(푋푛) = det⁡(푀푛). det(푀푛) det(푀푛) 퐷푛 This can be derived into det⁡(푋푛) = , and since det(푋푛) = 푥푛, 푥푛 = = . det(퐴) det(퐴) 퐷

And here ends the derivation of Cramer’s Rule, a method of solving system that is often seen as ineffective. Below is an example of Cramer’s Rule.

푥 − 푦 + 푧 = 5 EX: Solve the system: { 푥 + 2푦 − 4푧 = 6 . 2푥 − 3푦 + 푧 = −1 1 −1 1 5 This system forms the augmented matrix [1 2 −4| 2 ], which indicates D = −8. 2 −3 1 −1 Meanwhile, 퐷푥 = −64, 퐷푦 = −56, and 퐷푧 = −32. −64 푥 = = 8 −8 −56 Therefore, the solution is 푦 = = 7. −8 −32 푧 = = 4 { −8

Practice Questions (No CALC)

Part I. Calculate the determinants of following matrices: *det(퐴) × det(퐵) = det⁡(퐴퐵), this will help you bit more than other properties of determinants can, but other properties can help you as well. Matrix Determinant Matrix Determinant 2 3 4 2 (1) [ ] 7 (2) [ ] 12 1 5 8 7 9 10 1 6 (3) [ ] 6 (4) [ ] 33 3 4 −4 9 4 −6 −5 −4 (5) [ ] 0 (6) [ ] 39 −2 3 1 −7 10 0 12 12 (7) [ ] 200 (8) [ ] 336 5 20 −20 8 42 49 28 40 (9) [ ] 1519 (10) [ ] 2896 35 77 −64 12 −132 −99 13 15 (11) [ ] −4719 (12) [ ] 0 121 55 0 0 −4 2 300 −15 (13) [ ] 16 (14) [ ] −31725 −10 1 −15 −105 3 2 1 0 2 4 1 93 (15) [2 4 6] (16) [5 3 7] 9 6 3 5 2 1 6 3 8 −56 1 5 3 −36 (17) [4 8 2] (18) [2 3 1] 6 2 7 4 3 5 10 −2 5 −312 18 9 −3 −216 (19) [−9 3 3 ] (20) [ 0 6 −12] 6 2 −3 −12 −6 0 −18 9 0 −1620 5 −2 9 −9 (21) [ 3 6 9 ] (22) [−3 5 −7] 0 0 12 3 1 4 64 16 12 −39808 75 15 90 16875 (23) [56 0 40] (24) [105 135 15] 36 20 28 30 45 0 0 1 1 2 145 156 167 0 (25) [1 0 1] (26) [ 0 0 0 ] 1 1 0 287 276 265 2 4 6 8 −5824 3 15 9 27 98415 12 10 4 16 6 15 18 21 (27) [ ] (28) [ ] 12 20 12 16 0 30 42 15 18 14 0 2 39 0 0 12 1 1 1 2 2 10 10 10 20 12 1 1 2 2 10 10 20 20 (29) [ ] (30) [ ] 1 2 2 2 10 20 20 20 2 2 2 2 20 20 20 20

Part II. Find the inverse matrix of following matrices with (a) Gauss-Jordan Method and (b) determinant and adjugate matrices. Later, (c) reflect on the efficiency of both methods on finding inverse matrices of different size matrices. Matrix Inverse Matrix Inverse 2 3 5 3 4 2 7 1 (1) [ ] (2) [ ] 1 5 − 8 7 − [ 7 7] [ 12 6] 1 2 2 1 − − 7 7 3 3 9 10 1 6 3 2 (3) [ ] 2 5 (4) [ ] 3 4 − −4 9 − [ 3 3] [11 11] 1 3 4 1 − 2 2 33 33 4 −6 −5 −4 7 4 (5) [ ] 푁표⁡푖푛푣푒푟푠푒 (6) [ ] −2 3 1 −7 − [ 39 39 ] 1 5 − − 39 39 10 0 1 12 12 1 1 (7) [ ] (8) [ ] 5 20 0 −20 8 − [ 10 ] [42 28] 1 1 5 1 − 40 20 84 28 42 49 11 1 28 40 3 5 (9) [ ] (10) [ ] 35 77 − −64 12 − [ 217 31] [724 362] 5 6 4 7 − 217 217 181 724 −132 −99 13 15 (11) [ ] 5 3 (12) [ ] 푁표⁡푖푛푣푒푟푠푒 121 55 0 0 [ 429 143 ] 1 4 − − 39 143 −4 2 1 1 300 −15 7 1 (13) [ ] (14) [ ] −10 1 − −15 −105 1 [16 8] [2085 2085] 5 1 2085 1 4 − 8 4 2085 417 3 2 1 푁표⁡푖푛푣푒푟푠푒 2 4 1 11 2 25 (15) [2 4 6] (16) [5 3 7] − − 93 93 93 9 6 3 5 2 1 10 1 3 − − 31 31 31 5 16 14 − − [ 93 93 93] 6 3 8 13 5 29 1 5 3 3 4 1 (17) [4 8 2] − (18) [2 3 1] − 14 56 28 36 9 9 6 2 7 2 3 5 4 3 5 1 7 5 − − 7 28 14 6 36 36 5 3 9 1 17 7 − − − [ 7 28 14] [ 6 36 36 ] (19) 5 1 7 (20) 1 1 5 10 −2 5 − 18 9 −3 − 104 78 104 3 12 12 [−9 3 3 ] 3 5 25 [ 0 6 −12] 2 1 − −1⁡ 6 2 −3 104 26 104 −12 −6 0 3 6 3 4 1 1 1 − − 0 − [ 26 39 26] [ 3 2] (21) 2 1 1 (22) 3 1 59 −18 9 0 − − 5 −2 9 − − 45 15 20 5 45 45 [ 3 6 9 ] 1 2 1 [−3 5 −7] 1 7 8 − − 0 0 12 45 15 10 3 1 4 5 45 45 1 2 1 31 0 0 − − [ 12 ] [ 5 45 45] 64 16 12 25 13 5 (24) 1 6 53 (23) [56 0 40] − 75 15 90 − − 1244 2488 311 25 25 75 36 20 28 1 85 59 [105 135 15] 2 4 37 − − 311 2488 1244 30 45 0 75 25 75 35 11 7 1 13 38 − − [ 1244 622 311 ] [ 25 75 75 ] 0 1 1 1 1 1 45 56 67 푁표⁡푖푛푣푒푟푠푒 (25) [1 0 1] − (26) [ 0 0 0 ] 2 2 2 1 1 0 1 1 1 87 76 65 − 2 2 2 1 1 1 − [ 2 2 2] 107 9 101 12 8 32 4 23 (27) − − (28) − 2 4 6 8 364 364 728 91 3 15 9 27 729 729 243 729 12 10 4 16 33 3 16 8 6 15 18 21 772 1387 143 154 [ ] − − [ ] − 12 20 12 16 0 30 42 15 91 182 91 91 3645 3645 1215 3645 18 14 0 2 165 53 69 10 39 0 0 12 101 161 10 17 − − − − − 364 364 728 91 729 729 243 729 3 3 1 1 26 104 13 14 − − − − − [ 28 28 56 14] [ 729 729 243 729] 1 1 1 2 0 0 −1 1 1 1 (30) 0 0 − 1 1 2 2 0 −1 1 0 10 10 10 20 (29) [ ] 10 10 1 2 2 2 −1 1 0 0 10 10 20 20 1 1 [ ] 0 − 0 2 2 2 2 1 10 20 20 20 1 0 0 − 10 10 [ 2] 20 20 20 20 1 1 − 0 0 10 10 1 1 0 0 − [ 10 20]

Part III. Solve the following systems of equations with (a) augmented matrices, (b) inverse matrices, and (c) Cramer’s Rule. System Solutions System Solutions 푒푞1: 2푥 + 푦 = 4 푥 = 1 푒푞1: 3푥 + 4푦 = 13 푥 = 3 (1) { { (2) { { 푒푞2: 2푥 − 푦 = 0 푦 = 2 푒푞2: 6푥 + 푦 = 19 푦 = 1 푒푞1: 푥 + 4 = 2푦 푥 = 6 푒푞1: 푥 + 푦 = 10 푥 = 3 (3) { { (4) { { 푒푞2: 푥 − 3푦 = −9 푦 = 5 푒푞2: 푦 − 푥 = 4 푦 = 7 푒푞1: 2푥 − 푦 = 2 푥 = 8 푒푞1: 5푥 + 9 = 푦 푥 = 14 (5) { { (6) { { 푒푞2: 푥 + 푦 = 22 푦 = 14 푒푞2: 푦 − 푥 = 65 푦 = 79 푒푞1: 푥 + 푦 = 17 푥 = 12 푒푞1: 푦 − 푥 = 27 푥 = 18 (7) { { (8) { { 푒푞2: 5푥 + 12푦 = 120 푦 = 5 푒푞2: 0.2푦 − 0.5푥 = 0 푦 = 45 푒푞1: 푥 + 9 = 푦 푥 = 4 푒푞1: 3푥 − 2푦 = 0 푥 = 38 (9) { { (10) { { 푒푞2: 푥 − 2푦 = −22 푦 = 13 푒푞2: 푦 − 푥 = 0.5푥 푦 = 57 푒푞1: 3푥 − 푦 + 2푧 = 7 푥 = 2 푒푞1: −푥 + 푦 − 푧 = 6 푥 = −4 (11) {푒푞2: 2푥 + 3푦 − 푧 = 11 {푦 = 3 (12) {푒푞2: 푥 − 4푦 + 2푧 = −22 { 푦 = 7 푒푞3: 푥 + 푦 + 푧 = 7 푧 = 2 푒푞3: 4푥 + 3푦 − 2푧 = −5 푧 = 5 푒푞1: 푥 − 푦 + 푧 = 13 푥 = 3 푒푞1: 푥 − 푦 + 푧 = 4 푥 = 3 (13) {푒푞2: 6푥 − 푦 − 푧 = 10 {푦 = −1 (14) { 푒푞2: 3푥 + 푦 + 푧 = 0 {푦 = −4 푒푞3: −9푦 − 푧 = 0 푧 = 9 푒푞3: 2푥 + 푦 + 푧 = −3 푧 = −5 푒푞1: 푥 − 푦 − 푧 = 0 푥 = −2 푒푞1: 푥 + 푦 − 푧 = 1 푥 = 11 (15) { 푒푞2: 푥 + 푦 = −9 {푦 = −7 (16) {푒푞2: 푥 − 푦 + 3푧 = 49 {푦 = −1 푒푞3: 푥 − 푧 = −7 푧 = 5 푒푞3: 2푥 + 푦 + 푧 = 30 푧 = 9 푒푞1: 2푥 + 푦 + 푧 = 13 푥 = −4 푒푞1: 푥 + 푦 + 2푧 = 0 푥 = 4 (17) {푒푞2: 3푥 − 푦 + 3푧 = −1 {푦 = 13 (18) { 푒푞2: 푥 − 푦 − 푧 = 5 { 푦 = 2 푒푞3: 푥 + 푦 − 푧 = 1 푧 = 8 푒푞3: 4푥 − 8푦 − 푧 = 3 푧 = −3 푒푞1: 2푥 − 푦 + 푧 = 0 푥 = 1 푒푞1: 2푥 + 푦 − 푧 = 7 푥 = 4 (19) {푒푞2: 7푥 − 푦 − 푧 = −5 {푦 = 7 (20) {푒푞2: 4푥 − 2푦 + 푧 = 9 {푦 = 8 푒푞3: 푥 + 푦 + 2푧 = 18 푧 = 5 푒푞3: 2푥 − 푦 + 푧 = 13 푧 = 9