How Can You Represent the System of Equations With a Matrix 11x-12y=7

German language mathematician Carl Friedrich Gauss (1777–1855).

Carl Friedrich Gauss lived during the late 18th century and early 19th century, just he is still considered one of the nigh prolific mathematicians in history. His contributions to the scientific discipline of mathematics and physics bridge fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the mode mathematicians have worked for the last two centuries.

In this section, we will learn Gaussian emptying for solving systems, using matrices.

Writing the Augmented Matrix of a System of Equations

A matrix can serve as a device for representing and solving a system of equations. To express a organisation in matrix grade, nosotros excerpt the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this class, we phone call information technology an augmented matrix.

For example, consider the post-obit two × 2system of equations.

3x + 4y = 7

410 − twoy = 5

We can write this organisation as an augmented matrix (delight note that a more common formatting for augmented matrices has a solid vertical line running through the matrix, rather than a line on each row):

[latex]\displaystyle{\left[\matrix{{iii}&{4}&{|}{seven}\\{4}&-{2}&{|}{5}}\right]}[/latex]

We can besides write a matrix containing only the coefficients. This is called the
coefficient matrix.

[latex]\displaystyle{\left[\matrix{{3}&{4}\\{4}&-{2}}\correct]}[/latex]

A three-by-3 system of equations such every bit

iiix − y − z = 0

x + y = 5

2x − 3z = 2

has a coefficient matrix

[latex]\displaystyle{\left[\matrix{{3}&-{1}&-{1}\\{1}&{1}&{0}\\{2}&{0}&-{three}}\right]}[/latex]

and is represented by the augmented matrix
[latex]\displaystyle{\left[\matrix{{iii}&-{1}&-{one}&{\mid}&{0}\\{one}&{1}&{0}&{\mid}&{5}\\{2}&{0}&-{3}&{\mid}&{2}}\correct]}[/latex]

Discover that the matrix is written so that the variables line up in their own columns:
x-terms go in the commencement column, y-terms in the second column, and z-terms in the third column. Information technology is very important that each equation is written ax + past + cz = d in standard form so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.

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How To

Given a system of equations, write an augmented matrix.

1. Write the coefficients of the x-terms as the numbers down the offset column.
2. Write the coefficients of the y-terms as the numbers down the second cavalcade.
3. If there are z-terms, write the coefficients as the numbers downwards the third cavalcade.
4. Draw a vertical line and write the constants to the correct of the line.

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Example i

1. Write the augmented matrix for the given organization of equations.

x + 2 y − z = 3
2 x − y + two z = 6
x − 3 y + iii z = 4

2.  Write the augmented matrix of the given system of equations.

4 x − 3 y = 11

3 ten + 2 y = iv

Solutions

1. The augmented matrix displays the coefficients of the variables, and an additional column for the constants.
   [latex]\displaystyle{\left[\matrix{{1}&{ii}&-{1}&{\mid}&{3}\\{2}&-{1}&{two}&{\mid}&{6}\\{one}&-{3}&{3}&{\mid}&{4}}\right]}[/latex]
2. [latex]\displaystyle{\left[\matrix{{4}&-{three}&{\mid}&{11}\\{3}&{two}&{\mid}&{iv}}\right]}[/latex]

Writing a Organization of Equations from an Augmented Matrix

We can use augmented matrices to help united states of america solve systems of equations because they simplify operations when the systems are non encumbered by the variables. However, it is important to empathize how to move back and along betwixt formats in club to brand finding solutions smoother and more intuitive. Hither, nosotros will employ the information in an augmented matrix to write the arrangement of equations in standard class.

Instance two

1. Find the system of equations from the augmented matrix.
   [latex]\displaystyle{\left[\matrix{{1}&-{iii}&-{5}&{\mid}&-{two}\\{2}&-{5}&-{4}&{\mid}&{5}\\-{3}&{five}&{4}&{\mid}&{6}}\correct]}[/latex]
2. Write the system of equations from the augmented matrix.
   [latex]\displaystyle{\left[\matrix{{1}&-{1}&{one}&{\mid}&{v}\\{ii}&-{1}&{three}&{\mid}&{ane}\\{0}&{1}&{i}&{\mid}&-{nine}}\correct]}[/latex]

Solutions:

1) When the columns represent the variables            x,          y          ,          and          z,

x-3y-5z=-2

2x-5y-4z=5

-3x+5y+4z=6

2) When the columns correspond the variables            ten,          y          ,          and          z,

Performing Row Operations on a Matrix

At present that we can write systems of equations in augmented matrix form, nosotros will examine the diverserow operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows.

Performing row operations on a matrix is the method nosotros use for solving a system of equations. In social club to solve the system of equations, nosotros want to catechumen the matrix to
row-echelon form, in which at that place are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position beneath the main diagonal as shown.

Row-echelon form:

[latex]\displaystyle{\left[\matrix{{1}&{a}&{b}\\{0}&{one}&{d}\\{0}&{0}&{1}}\correct]}[/latex]

Nosotros use row operations corresponding to equation operations to obtain a new matrix that is
row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form.

1. In any nonzero row, the first nonzero number is a ane. It is called a leading ane.
2. Any all-naught rows are placed at the bottom on the matrix.
3. Whatever leading 1 is below and to the right of a previous leading one.
4. Any column containing a leading i has zeros in all other positions in the column.

To solve a organisation of equations nosotros tin can perform the following row operations to catechumen the
coefficient matrix to row-echelon course and do back-exchange to notice the solution.

1. Interchange rows. (Notation: [latex]\displaystyle{R}_{{i}}\leftrightarrow{R}_{{j}}[/latex])
2. Multiply a row by a abiding. (Notation: [latex]\displaystyle{c}{R}_{{i}}[/latex])
3. Add together the product of a row multiplied past a constant to another row. (Notation: [latex]\displaystyle{R}_{{i}}+{c}{R}_{{j}}[/latex])

Each of the row operations corresponds to the operations we accept already learned to solve systems of equations in iii variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a i as the first entry and then that row 1 can be used to catechumen the remaining rows.

Gaussian Elimination

TheGaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix Awith the number i as the entry downwards the main diagonal and accept all zeros below.

[latex]\displaystyle{A}={\left[\matrix{{a}_{{11}}&{a}_{{12}}&{a}_{{13}}\\{a}_{{21}}&{a}_{{22}}&{a}_{{23}}\\{a}_{{31}}&{a}_{{32}}&{a}_{{33}}}\right]}[/latex]→[latex]{\left[\matrix{{one}&{b}_{{12}}&{b}_{{13}}\\{0}&{i}&{b}_{{23}}\\{0}&{0}&{1}}\right]}[/latex]

The first footstep of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below.

Of import Notation: Nosotros Volition FOCUS ON GETTING OUT MATRIX IN REDUCED ROW ECHELON FORM WHICH Will Await LIKE THIS;

[latex]{\left[\matrix{{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}}\right]}[/latex]

SEE NEXT SECTION

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How To

Given an augmented matrix, perform row operations to achieve row-echelon form.

1. The commencement equation should accept a leading coefficient of 1. Interchange rows or multiply by a abiding, if necessary.
2. Employ row operations to obtain zeros down the first column below the kickoff entry of 1.
3. Utilise row operations to obtain a 1 in row two, cavalcade 2.
4. Use row operations to obtain zeros downwards column two, below the entry of i.
5. Use row operations to obtain a 1 in row 3, column 3.
6. Proceed this process for all rows until there is a i in every entry downwardly the main diagonal and in that location are but zeros below.
7. If any rows comprise all zeros, place them at the bottom.

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Example 3

i) Solve the given system by Gaussian elimination.

210 + 3y = 6
10 − y = [latex]\displaystyle\frac{{1}}{{ii}}[/latex]

2) Solve the given organization by Gaussian elimination.

410 + iiiy = eleven
10 − 3y = −1

3) Utilize Gaussian emptying to solve the given 2 × 2 arrangement of equations.

2x + y = 1
4x +2y=six

4) Solve the system of equations.
3 x + 4 y = 12
6 x + eight y = 24

five) Perform row operations on the given matrix to obtain row-echelon form.
[latex]\displaystyle{\left[\matrix{{1}&-{three}&{4}&{\mid}&{3}\\{ii}&-{five}&{six}&{\mid}&{6}\\-{3}&{3}&{4}&{\mid}&{vi}}\right]}[/latex]

6) Write the system of equations in row-echelon form.
[latex]\displaystyle{\left[\matrix{{ten}-{2}{y}+{iii}{z}={ix}\\-{10}+{three}{y}=-{4}\\{2}{x}-{5}{y}+{five}{z}={17}}\right]}[/latex]

Solutions

1. Starting time, we write this as an augmented matrix.
   [latex]\displaystyle{\left[\matrix{{2}&{iii}&{\mid}&{6}\\{1}&-{1}&{\mid}&\frac{{1}}{{2}}}\right]}[/latex]
   • Nosotros want a ane in row 1, cavalcade 1.
   • This tin be accomplished by interchanging row i and row 2.[latex]\displaystyle{R}_{{1}}\rightarrow{R}_{{two}}{\left[\matrix{{one}&-{1}&{\mid}&\frac{{1}}{{ii}}\\{ii}&{three}&{\mid}&{6}}\right]}[/latex]
2. We at present have a 1 every bit the first entry in row 1, column 1.
   • Now permit's obtain a 0 in row 2, cavalcade.
   • This can be achieved by multiplying row 1 by –two and then adding the result to row
   • [latex]\displaystyle-{ii}{R}_{{1}}+{R}_{{2}}={R}_{{ii}}\rightarrow{\left[\matrix{{1}&{1}&{\mid}&\frac{{ane}}{{2}}\\{0}&{5}&{\mid}&{v}}\right]}[/latex]
   • We merely take one more step, to multiply row two by  [latex]\displaystyle\frac{{1}}{{5}}[/latex]
   • Use dorsum-commutation. The second row of the matrix represents
     y = ane. Back-substitute y = i into the showtime equation.
     [latex]\displaystyle{\left[\matrix{{x}-{({1})}=\frac{{one}}{{2}}\\{ten}=\frac{{3}}{{2}}}\right]}[/latex]
   • The solution is the signal  [latex]\displaystyle{(\frac{{3}}{{2}},{1})}[/latex].
3. (i,2)
4. Write the system as an augmented matrix.
   • [latex]\displaystyle{\left[\matrix{{2}&{1}&{\mid}&{i}\\{four}&{2}&{\mid}&{6}}\right]}[/latex]
   • Obtain a 1 in row i, column one. This tin be accomplished by multiplying the first row by [latex]\displaystyle\frac{{1}}{{2}}[/latex]
   • Next, we want a 0 in row 2, column 1.
   • Multiply row one by –4 and add row ane to row 2.[latex]\displaystyle-{4}{R}_{{1}}+{R}_{{ii}}={R}_{{2}}\rightarrow{\left[\matrix{{one}&\frac{{1}}{{2}}&{\mid}&\frac{{one}}{{2}}\\{0}&{0}&{\mid}&{iv}}\right]}[/latex]
   • The second row represents the equation 0 = iv.
   • Therefore, the system is inconsistent and has no solution.
5. Perform row operations on the augmented matrix to endeavour and accomplish row-echelon grade .
   • [latex]\displaystyle{A}={\left[\matrix{{3}&{4}&{\mid}&{12}\\{half dozen}&{8}&{\mid}&{24}}\right]}[/latex]
   • The matrix ends upwardly with all zeros in the terminal row: 0 y = 0. Thus, there are an infinite number of solutions and the organisation is classified every bit dependent. To discover the generic solution, return to one of the original equations and solve for y
   • [latex]\displaystyle{\left[\matrix{{3}{ten}+{4}{y}={12}\\{4}{y}={12}-{three}{x}\\{y}={3}-\frac{{iii}}{{4}}{x}}\right]}[/latex]
   • So the solution to this system is [latex]\displaystyle{({x},{3}-\frac{{3}}{{4}}{x})}[/latex].
6. The first row already has a i in row 1, column ane. The next step is to multiply row 1 by –two and add it to row ii. Then supervene upon row 2 with the effect.
   • [latex]\displaystyle-{2}{R}_{{1}}+{R}_{{two}}={R}_{{2}}\rightarrow{\left[\matrix{{1}&-{3}&{four}&{\mid}&{3}\\{0}&{1}&-{2}&{\mid}&{0}\\-{3}&{3}&{4}&{\mid}&{6}}\right]}[/latex]
   • Next, obtain a zippo in row 3, cavalcade 1.[latex]\displaystyle{iii}{R}_{{i}}+{R}_{{3}}={R}_{{3}}\rightarrow{\left[\matrix{{1}&-{iii}&{iv}&{\mid}&{3}\\{0}&{1}&-{2}&{\mid}&{0}\\{0}&-{6}&{16}&{\mid}&{fifteen}}\right]}[/latex]
   • Next, obtain a zero in row three, cavalcade ii.
   • [latex]\displaystyle{6}{R}_{{2}}+{R}_{{3}}={R}_{{three}}\rightarrow{\left[\matrix{{1}&-{3}&{4}&{\mid}&{3}\\{0}&{1}&-{2}&{\mid}&{0}\\{0}&{0}&{4}&{\mid}&{15}}\right]}[/latex]
   • The last step is to obtain a one in row 3, column 3.
   • [latex]\displaystyle\frac{{1}}{{2}}{R}_{{3}}={R}_{{3}}\rightarrow{\left[\matrix{{1}&-{3}&{iv}&{\mid}&{iii}\\{0}&{1}&-{two}&{\mid}&-{6}\\{0}&{0}&{1}&{\mid}&\frac{{21}}{{2}}}\right]}[/latex]
7. [latex]\displaystyle{\left[\matrix{{1}&-\frac{{5}}{{two}}&\frac{{5}}{{ii}}&{\mid}&\frac{{17}}{{2}}\\{0}&{1}&{5}&{\mid}&{9}\\{0}&{0}&{i}&{\mid}&{2}}\right]}[/latex]

Solving a Organization of Linear Equations Using Matrices

We accept seen how to write a system of equations with an augmented matrix , and and so how to use row operations and back-substitution to obtain row-echelon form . Now, we will take row-echelon form a pace farther to solve a 3 by 3 system of linear equations. The full general thought is to eliminate all but one variable using row operations and so back-substitute to solve for the other variables.

Case 4

1. Solve the system of linear equations using matrices.
   ten − y + z = 8
   2 x + 3 y − z = −2
   3 x − 2 y − ix z = nine
2. Solve the post-obit organization of linear equations using matrices.
   − x − 2 y + z = −1
   2 x + 3 y = 2
   y − 2 z = 0
3. Solve the system using matrices.
   ten + 4 y − z = 4
   2 x + v y + 8 z = 15
   ten + 3 y − iii z = 1

Solutions

1. Beginning, nosotros write the augmented matrix.
   [latex]\displaystyle{\left[\matrix{{ane}&-{1}&{i}&{\mid}&{8}\\{2}&{3}&-{ane}&{\mid}&-{two}\\{three}&-{2}&-{9}&{\mid}&{nine}}\right]}[/latex]Adjacent, we perform row operations to obtain row-echelon form.[latex]\displaystyle-{two}{R}_{{1}}+{R}_{{2}}={R}_{{two}}\rightarrow{\left[\matrix{{1}&-{one}&{1}&{\mid}&{8}\\{0}&{v}&-{3}&{\mid}&-{eighteen}\\{3}&-{2}&-{nine}&{\mid}&{9}}\right]}[/latex]The easiest fashion to obtain a 1 in row 2 of cavalcade ane is to interchange
   R _ii and R _three.
   [latex]\displaystyle{R}_{{2}}\rightarrow{R}_{{3}}\rightarrow{\left[\matrix{{i}&-{ane}&{1}&{\mid}&{8}\\{0}&{1}&-{12}&{\mid}&-{15}\\{0}&{5}&-{three}&{\mid}&-{18}}\right]}[/latex]The concluding matrix represents the equivalent system.[latex]\displaystyle{\left[\matrix{{x}-{y}+{z}={8}\\{y}-{12}{z}=-{15}\\{z}={1}}\right]}[/latex]Using back-substitution, we obtain the solution as [latex]\displaystyle{({iv},-{3},{1})}[/latex].
2. Write the augmented matrix, [latex]\displaystyle{\left[\matrix{-{one}&-{two}&{1}&{\mid}&-{i}\\{2}&{3}&{0}&{\mid}&{2}\\{0}&{i}&-{two}&{\mid}&{0}}\right]}[/latex]First, multiply row ane past –1 to get a ane in row 1, column ane. Then, perform row operations to obtain row-echelon form. [latex]\displaystyle-{R}_{{ane}}\rightarrow{\left[\matrix{{1}&{2}&-{i}&{\mid}&{1}\\{ii}&{3}&{0}&{\mid}&{2}\\{0}&{1}&-{two}&{\mid}&{0}}\correct]}[/latex]The last matrix represents the following organisation.

ten + twoy − z = 1
y − 2z = 0
0 = 0

Nosotros see past the identity 0 = 0, that this is a dependent system with an space number of solutions. We then detect the generic solution. By solving the second equation for y and substituting it into the first equation we tin can solve forz in terms of 10.

[latex]\displaystyle{{x}+{2}{y}-{z}={ane}, {y}={2}{z}\\{10}+{2}{\left({2}{z}\right)}-{z}={ane}}\\{{x}+{iii}{z}={1}}\\{3z}={1}-{x}\\{z}=\frac{{{1}-{ten}}}{{3}}[/latex]

Now nosotros substitute the expression forz into the second equation to solve for y in terms of ten.

[latex]\displaystyle{{y}-{two}{z}={0}, {z}=\frac{{{1}-{x}}}{{three}}\\{y}-{ii}{\left(\frac{{{1}-{10}}}{{3}}\right)}={0}\\{y}=\frac{{{2}-{2}{ten}}}{{three}} ={0}}[/latex]

The generic solution is [latex]\displaystyle{({x},\frac{{{2}-{2}{x}}}{{three}},\frac{{{1}-{x}}}{{3}})}[/latex].

3.  (i,1,1)

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Q&A

Tin any system of linear equations be solved past Gaussian elimination?

Yes, a system of linear equations of any size tin can be solved by Gaussian elimination.

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How To

Given a organization of equations, solve with matrices using a calculator.

1. Salvage the augmented matrix as a matrix variable [A], [B ], [C ], . . .
2. Use the ref( role in the calculator, calling up each matrix variable equally needed.

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Instance 5

one)Solve the system of equations.5
ten + 3y + 9z = −1 −2
10 + 3y − z = −2 −
x − 4y + fivez = 1

ii)Carolyn invests a full of $12,000 in two municipal bonds, ane paying x.five% interest and the other paying 12% involvement. The almanac interest earned on the 2 investments last year was $one,335. How much was invested at each rate?

3)Ava invests a full of $10,000 in iii accounts, one paying 5% interest, another paying viii% interest, and the tertiary paying 9% interest. The annual interest earned on the three investments terminal year was $770. The amount invested at 9% was twice the corporeality invested at five%. How much was invested at each rate?

four)A modest shoe company took out a loan of $one,500,000 to aggrandize their inventory. Part of the coin was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at x% was four times the amount borrowed at seven%, and the annual interest on all iii loans was $130,500. Employ matrices to notice the amount borrowed at each rate.

Solutions

1. Write the augmented matrix for the system of equations.
   [latex]\displaystyle{\left[\matrix{{5}&{3}&{9}&{\mid}&-{i}\\-{ii}&{3}&-{1}&{\mid}&-{2}\\-{ane}&-{four}&{v}&{\mid}&-{i}}\right]}[/latex]On the matrix page of the calculator, enter the augmented matrix above as the matrix variable [A]
   [latex]\displaystyle{[{A}]}={\left[\matrix{{5}&{3}&{9}&{\mid}&-{i}\\-{2}&{three}&-{1}&{\mid}&-{ii}\\-{1}&-{4}&{five}&{\mid}&-{i}}\right]}[/latex]
   
   Utilize theref( role in the computer, calling upwardly the matrix variable )
   Evaluate using back-exchange. The solution is
   [latex]\displaystyle{(\frac{{61}}{{187}},-\frac{{92}}{{187}},-\frac{{24}}{{187}})}[/latex]. (If yous use the rref functions it will give you lot the answer without back substitution.)
2. We accept a system of two equations in two variables. Permit ten = the corporeality invested at ten.5% involvement, and y = the amount invested at 12% involvement.
   [latex]\displaystyle{\left[\matrix{{x}+{y}={12000}\\{0.105}{x}+{0.12}{y}={1335}}\right]}[/latex]Equally a matrix, we accept[latex]\displaystyle{\left[\matrix{{1}&{i}&{\mid}&{12000}\\{0.015}&{0.12}&{\mid}&{1335}}\right]}[/latex]Multiply row 1 by –0.015 and add the upshot to row 2.[latex]\displaystyle{\left[\matrix{{1}&{one}&{\mid}&{12000}\\{0}&{0.015}&{\mid}&{75}}\right]}[/latex]Then,0.015
   y = 75
   y = 5,000So 12000 – 5000=7000.Thus, $five,000 was invested at 12% interest and $7,000 at x.v% interest.
3. We accept a system of three equations in three variables. Let x exist the corporeality invested at 5% interest, permit y exist the amount invested at viii% interest, and permit z be the amount invested at nine% interest. Thus,
   x + y + z = 10 , 000
   0.05 x + 0.08 y + 0.09 z = 770
   2 x − z = 0
   Equally a matrix, we accept
   [latex]\displaystyle{\left[\matrix{{1}&{i}&{i}&{\mid}&{10000}\\{0.05}&{0.08}&{0.09}&{\mid}&{770}\\{ii}&{0}&-{1}&{\mid}&{0}}\right]}[/latex]At present, we perform Gaussian emptying to achieve row-echelon form.[latex]\displaystyle-{0.05}{R}_{{1}}+{R}_{{ii}}={R}_{{2}}\rightarrow{\left[\matrix{{one}&{1}&{1}&{\mid}&{10000}\\{0}&{0.03}&{0.04}&{\mid}&{270}\\{2}&{0}&-{1}&{\mid}&{0}}\correct]}[/latex].  The third row tells usa
   [latex]\displaystyle{\frac{-{i}}{{3}}{z}=-{2000}}[/latex];
   thus,z = 6000. The second row tells us [latex]\displaystyle{y}+{\frac{{4}}{{3}}{z}={9000}}[/latex]. Substituting z = 6000, we get[latex]\displaystyle{\left[\matrix{{y}+{\frac{{4}}{{3}}{({6000})}={9000}\\{y}+{8000}={9000}\\{y}={g}}}\right]}[/latex]The commencement row tells u.s.a.
   x + y + z = 10000. Substituting y = thou and z = 6000, nosotros get
   ten + 1 , 000 + vi , 000 = x,000; x = iii,000.
   The answer is $3,000 invested at 5% interest, $1,000 invested at viii%, and $6,000 invested at nine% involvement.
4. $150,000 at seven%, $750,000 at viii%, $600,000 at ten%

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Key Concepts

• An augmented matrix is ane that contains the coefficients and constants of a system of equations.
• A matrix augmented with the constant column can exist represented equally the original system of equations.
• Row operations include multiplying a row by a abiding, adding i row to some other row, and interchanging rows.
• We can use Gaussian elimination to solve a system of equations.
• Row operations are performed on matrices to obtain row-echelon course.
• To solve a organization of equations, write it in augmented matrix class. Perform row operations to obtain row-echelon form. Back-substitute to find the solutions.
• A calculator can be used to solve systems of equations using matrices.
• Many real-world problems can exist solved using augmented matrices.

Media

Access these online resources for additional education and practice with solving systems of linear equations using Gaussian elimination.

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OpenStax, Precalculus, "Solving Systems with Gaussian Elimination," licensed under a CC BY 3.0 license.

Mathispower4u, "
Ex 2: Solve a System of 2 Equations with Using an Augmented Matrix (Row Echelon Class)," licensed under a Standard YouTube license.

Mathispower4u, "Ex 2: Solve a System of Three Equations with Using an Augmented Matrix (Row Echelon Form)," licensed nether a Standard YouTube license.

Mathispower4u, "
Augmented Matrices on the TI83/84," licensed under a Standard YouTube license.

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Source: https://courses.lumenlearning.com/sanjacinto-finitemath1/chapter/reading-solving-systems-with-gaussian-elimination/

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