Lesson Solving systems of linear equations in three unknowns using determinant (Cramer's rule)

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<H2>Solving systems of linear equations in three unknowns using determinant (Cramer's rule)</H2>

In this lesson you will find typical examples on solution the systems of three linear equations in three unknowns using determinant &nbsp;(Cramer's rule).

The lesson is a continuation of the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/HOW-TO-solve-system-of-linear-eqns-in-three-unknowns-using-det.lesson>HOW TO solve system of linear equations in three unknowns using determinant (Cramer's rule)</A>&nbsp; under the current topic in this site.

<H3>Cramer's rule</H3>Let

{{{system(a*x + b*y + c*z = u, 
d*x + e*y + f*z = v, 
g*x + h*y + k*z = w)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

be a system of three linear equations in three unknowns, where &nbsp;{{{a}}}, &nbsp;{{{b}}}, &nbsp;{{{c}}}, &nbsp;{{{d}}}, &nbsp;{{{e}}}, &nbsp;{{{f}}}, &nbsp;{{{g}}}, &nbsp;{{{h}}}, {{{k}}}&nbsp; are coefficients; &nbsp;{{{u}}}, &nbsp;{{{v}}}&nbsp; and &nbsp;{{{w}}}&nbsp; are right side constants; &nbsp;{{{x}}}, &nbsp;{{{y}}}&nbsp; and &nbsp;{{{z}}}&nbsp; are unknowns. &nbsp;To solve the system of linear equations &nbsp;(1)&nbsp; using the &nbsp;Cramer's rule&nbsp; you need

1. &nbsp;to form the coefficient matrix &nbsp;{{{(matrix(3,3, a,b,c,d,e,f,g,h,k))}}}&nbsp; and calculate its determinant &nbsp;D = det &nbsp;{{{(matrix(3,3, a,b,c,d,e,f,g,h,k))}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;If the determinant is not zero, &nbsp;then the solution does exist and is unique, &nbsp;and the following steps are applicable.

2. &nbsp;to calculate the &nbsp;<U>first unknown</U>&nbsp; you need to modify the coefficient matrix replacing its &nbsp;<U>first column</U>&nbsp; by the <U>right side column</U>. &nbsp;You get the matrix &nbsp;{{{(matrix(3,3, u,b,c,v,e,f,w,h,k))}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Then calculate the determinant of this modified matrix &nbsp;Dx = det{{{(matrix(3,3, u,b,c,v,e,f,w,h,k))}}}&nbsp; and divide it by the determinant of the coefficient matrix: &nbsp;{{{x}}} = {{{Dx/D}}}.

3. &nbsp;to calculate the &nbsp;<U>second unknown</U>&nbsp; you need to modify the coefficient matrix replacing its &nbsp;<U>second column</U>&nbsp; by the <U>right side column</U>. &nbsp;You get the matrix &nbsp;{{{(matrix(3,3, a,u,c,d,v,f,g,w,k))}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Then calculate the determinant of this modified matrix &nbsp;Dy = det{{{(matrix(3,3, a,u,c,d,v,f,g,w,k))}}}&nbsp; and divide it by the determinant of the coefficient matrix: &nbsp;{{{y}}} = {{{Dy/D}}}.

4. &nbsp;to calculate the &nbsp;<U>third unknown</U>&nbsp; you need to modify the coefficient matrix replacing its &nbsp;<U>third column</U>&nbsp; by the <U>right side column</U>. &nbsp;You get the matrix &nbsp;{{{(matrix(3,3, a,b,u,d,e,v,g,h,w))}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Then calculate the determinant of this modified matrix &nbsp;Dy = det{{{(matrix(3,3, a,b,u,d,e,v,g,h,w))}}}&nbsp; and divide it by the determinant of the coefficient matrix: &nbsp;{{{y}}} = {{{Dy/D}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;At this stage the solution is completed.

Thus &nbsp;&nbsp;&nbsp;&nbsp;{{{x = det(matrix(3,3,u,b,c,v,e,f,w,h,k))/det(matrix(3,3,a,b,c,d,e,f,g,h,k))}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{y = det(matrix(3,3,a,u,c,d,v,f,g,w,k))/det(matrix(3,3,a,b,c,d,e,f,g,h,k))}}},&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{z = det(matrix(3,3,a,b,u,d,e,v,g,h,w))/det(matrix(3,3,a,b,c,d,e,f,g,h,k))}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)

<H3>Example 1</H3>Solve the system using the Cramer's rule

{{{system(2x + 2y - z = 2,
x + y + z = 4,
x - y - z = 2)}}}.

<B>Solution</B>

The coefficient matrix is &nbsp;&nbsp;{{{D}}} = {{{(matrix(3,3, 2, 2, -1, 1, 1, 1, 1, -1, -1))}}}.

Let us calculate its determinant using co-factoring &nbsp;(expanding)&nbsp; along the first row &nbsp;(see the lesson &nbsp;<A HREF=>Co-factoring the determinant of a 3x3 matrix</A>&nbsp; under the current topic):

{{{det(D)}}} = 2*det{{{(matrix(2,2, 1,1,-1,-1))}}} - 2*det{{{(matrix(2,2, 1,1,1,-1))}}} + (-1)*det{{{(matrix(2,2, 1,1,1,-1))}}} = 2*(-1-(-1)) - 2*(-1-1) + (-1)*(-1-1) = 0 - 2*(-2) + (-1)*(-2) = 0 + 4 + 2 = 6.

The determinant is non-zero, &nbsp;hence, &nbsp;the Cramer's rule is applicable.

Now, &nbsp;let us calculate the determinant of the first modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{x}}}. &nbsp;Again, &nbsp;use the determinant expanding along the first row:

{{{Dx}}} = det{{{(matrix(3,3, 2, 2, -1, 4, 1, 1, 2, -1, -1))}}} = 2*det{{{(matrix(2,2, 1,1,-1,-1))}}} - 2*det{{{(matrix(2,2, 4,1,2,-1))}}} + (-1)*det{{{(matrix(2,2, 4,1,2,-1))}}} = 2*(-1-(-1)) - 2*(-4-2) + (-1)*(-4 -2) = 0 - 2*(-6) - 1*(-6) = 0 + 12 + 6 = 18.

Next, &nbsp;calculate the determinant of the second modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{y}}}. &nbsp;Use the determinant expanding along the first row:

{{{Dy}}} = det{{{(matrix(3,3, 2, 2, -1, 1, 4, 1, 1, 2, -1))}}} = 2*det{{{(matrix(2,2,  4,1,2,-1))}}} - 2*det{{{(matrix(2,2, 1,1,1,-1))}}} + (-1)*det{{{(matrix(2,2, 1,4,1,2))}}} = 2*(-4-2) - 2*(-1-1) + (-1)*(2-4) = 2*(-6) - 2*(-2) + (-1)*(-2) = -12 +4 + 2 = -6.

At last, &nbsp;calculate the determinant of the third modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{z}}}. &nbsp;Use the determinant expanding along the first row:

{{{Dz}}} = det{{{(matrix(3,3, 2, 2, 2, 1, 1, 4, 1, -1, 2))}}} = 2*det{{{(matrix(2,2,  1,4,-1,2))}}} - 2*det{{{(matrix(2,2, 1,4,1,2))}}} + 2*det{{{(matrix(2,2, 1,1,1,-1))}}} = 2*(2+4) - 2*(2-4) + 2*(-1-1) = 2*6 - 2*(-2) + 2*(-2) = 12.

Hence, &nbsp;{{{x}}} = {{{Dx/D}}} = {{{18/6}}} = {{{3}}}, &nbsp;{{{y}}} = {{{Dy/D}}} = {{{(-6)/6}}} = {{{-1}}}&nbsp; and &nbsp;{{{z}}} = {{{Dz/D}}} = {{{12/6}}} = {{{2}}}.

You can easily check this solution by substituting the found values into the original system of equations. &nbsp;The solution is unique.

<B>Answer</B>. &nbsp;{{{x}}} = {{{3}}}, &nbsp;{{{y}}} = {{{-1}}}&nbsp; and &nbsp;{{{z}}} = {{{2}}}.

<H3>Example 2</H3>Solve the system using the Cramer's rule

{{{system(y - 2z = -3,
x - y = 2,
3x + z = 11)}}}.

<B>Solution</B>

Let us rewrite the system using zero coefficients for variables that are not shown

{{{system(0*x + y - 2z = -3,
x - y + 0*z = 2,
3x +0*y + z = 11)}}}.

The coefficient matrix is &nbsp;&nbsp;{{{D}}} = {{{(matrix(3,3, 0, 1, -2, 1, -1, 0, 3, 0, 1))}}}.

{{{det(D)}}} = 0*det{{{(matrix(2,2, -1,0,0,1))}}} - 1*det{{{(matrix(2,2, 1,0,3,1))}}} + (-2)*det{{{(matrix(2,2, 1,-1,3,0))}}} = 0 - 1*(1-0) + (-2)*(0+3) = 0 - 1 - 6 = -7.

The determinant is non-zero, &nbsp;hence, &nbsp;the Cramer's rule is applicable.

{{{Dx}}} = det{{{(matrix(3,3,  -3, 1, -2, 2, -1, 0, 11, 0, 1))}}} = (-3)*det{{{(matrix(2,2, -1,0,0,1))}}} - 1*det{{{(matrix(2,2, 2,0,11,1))}}} + (-2)*det{{{(matrix(2,2, 2,-1,11,0))}}} = (-3)*(-1) - 1*2 + (-2)*11 = 3 - 2 - 22 = -21.

Next, &nbsp;calculate the determinant of the second modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{y}}}. &nbsp;Use the determinant expanding along the first row:

{{{Dy}}} = det{{{(matrix(3,3, 0, -3, -2, 1, 2, 0, 3, 11, 1))}}} = 0*det{{{(matrix(2,2,  2,0,11,1))}}} - (-3)*det{{{(matrix(2,2, 1,0,3,1))}}} + (-2)*det{{{(matrix(2,2, 1,2,3,11))}}} = 0 - (-3)*1 + (-2)*(11-6) = 0 + 3 + (-2)*5 = 3 - 10 = -7.

At last, &nbsp;calculate the determinant of the third modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{z}}}. &nbsp;Use the determinant expanding along the first row:

{{{Dz}}} = det{{{(matrix(3,3, 0, 1, -3, 1, -1, 2, 3, 0, 11))}}} = 0*det{{{(matrix(2,2,  -1,2,0,11))}}} - 1*det{{{(matrix(2,2, 1,2,3,11))}}} + (-3)*det{{{(matrix(2,2, 1,-1,3,0))}}} = 0 - 1*(11-6) + (-3)*3 = 0 - 5 - 9 = -14.

Hence, &nbsp;{{{x}}} = {{{Dx/D}}} = {{{(-21)/(-7)}}} = {{{3}}}, &nbsp;{{{y}}} = {{{Dy/D}}} = {{{(-7)/(-7)}}} = {{{1}}}&nbsp; and &nbsp;{{{z}}} = {{{Dz/D}}} = {{{(-14)/(-7)}}} = {{{2}}}.

You can easily check this solution by substituting the found values into the original system of equations. &nbsp;The solution is unique.

<B>Answer</B>. &nbsp;{{{x}}} = {{{3}}}, &nbsp;{{{y}}} = {{{1}}}&nbsp; and &nbsp;{{{z}}} = {{{2}}}.

<H3>Example 3</H3>Solve the system using the Cramer's rule

{{{system(x + y + z = 0,
x - y - 3z = 0,
x + y - 2z = 0)}}}.

<B>Solution</B>

Notice that all right side terms are zeroes in the given system of equations. &nbsp;Such systems with the zero right side terms are called &nbsp;<B><U>homogeneous</U></B>&nbsp; systems.

The coefficient matrix is &nbsp;&nbsp;{{{D}}} = {{{(matrix(3,3, 1, 1, 1, 1, -1, -3, 1, 1, -2))}}}.

{{{det(D)}}} = 1*det{{{(matrix(2,2, -1,-3,1,-2))}}} - 1*det{{{(matrix(2,2, 1,-3,1,-2))}}} + 1*det{{{(matrix(2,2, 1,-1,1,1))}}} = 1*((-1)*(-2)-(-3)*1) - 1*(1*(-2)-1*(-3)) + 1*(1-(-1)) = 5 - 1  + 2 = 6.

The determinant is non-zero, &nbsp;hence, &nbsp;the Cramer's rule is applicable.

Now, the first modified matrix has the first column consisting of zeroes. &nbsp;If you calculate its determinant using co-factoring along the first column, &nbsp;you will get all the expanding terms equal to zero. &nbsp;Hence, &nbsp;the first modified matrix has the zero determinant.

Similarly, &nbsp;the second and the third modified matrices also have the zero determinants.

It implies that the given system of equations has the zero solution, &nbsp;and this solution is unique.

<B>Answer</B>. &nbsp;{{{x}}} = {{{0}}}, &nbsp;{{{y}}} = {{{0}}}&nbsp; and &nbsp;{{{z}}} = {{{0}}}.

Based on this example, &nbsp;you cam make a general conclusion that a &nbsp;<B>homogeneous system of linear equations with non-zero determinant has only zero solution</B>.

To complete this lesson, &nbsp;I will show you that sometimes the Cramer's rule can help you to solve even &nbsp;<U>non-linear</U>&nbsp;(!) systems of equations!

&nbsp;&nbsp;&nbsp;&nbsp;. . . Of course, &nbsp;if the given non-linear system of equations can be reduced to the linear one after replacing the unknowns :)

Below is an example.

<H3>Example 4</H3>Solve the system of three non-linear equations in three unknowns

&nbsp;{{{1/x}}} + {{{1/y}}} - {{{1/z}}} = {{{4}}},

-{{{1/x}}} + {{{1/y}}} + {{{1/z}}} = {{{6}}},

&nbsp;{{{1/x}}} + {{{1/y}}} + {{{1/z}}} = {{{8}}}.

<B>Solution</B>

Let us introduce new variables &nbsp;{{{p}}} = {{{1/x}}}, &nbsp;{{{q}}} = {{{1/y}}}&nbsp; and &nbsp;{{{r}}} = {{{1/z}}}.

Then the given system of non-linear equations is reduced to the linear one:

{{{system (p + q - r = 4,
           -p + q + r = 6,
           p + q + r = 8
)}}}

The coefficient matrix is &nbsp;{{{(matrix(3,3, 1, 1, -1, -1, 1, 1, 1, 1, 1))}}}. &nbsp;Its determinant is &nbsp;1*1*1 + 1*1*1 + 1*(-1)*(-1) - 1*1*(-1) - 1*1*1 - (-1)*1*1 = 1 + 1 + 1 + 1 - 1 + 1 = 4.

The determinant is non-zero. &nbsp;Hence, &nbsp;the Cramer's rule is applicable.

Let us calculate the determinant of the first modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{p}}}. Use the determinant expanding along the first row:

{{{Dp}}} = det{{{(matrix(3,3,  4, 1, -1, 6, 1, 1, 8, 1, 1))}}} = 4*det{{{(matrix(2,2, 1,1,1,1))}}} - 1*det{{{(matrix(2,2, 6,1,8,1))}}} + (-1)*det{{{(matrix(2,2, 6,1,8,1))}}} = 4*(1-1) - 1*(6-8) - 1*(6-8) = 0 + 2 + 2 = 4.

Now, &nbsp;let us calculate the determinant of the second modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{q}}}. &nbsp;Use the determinant expanding along the first row:

{{{Dq}}} = det{{{(matrix(3,3,  1, 4, -1, -1, 6, 1, 1, 8, 1))}}} = 1*det{{{(matrix(2,2, 6,1,8,1))}}} - 4*det{{{(matrix(2,2, -1,1,1,1))}}} + (-1)*det{{{(matrix(2,2, -1,6,1,8))}}} = 1*(6-8) - 4*(-1-1) + (-1)*(-8-6) = -2 + 8 + 14 = 20.

At last, &nbsp;let us calculate the determinant of the third modified matrix, &nbsp;which is the numerator of the fraction for &nbsp;{{{r}}}. &nbsp;Again, &nbsp;use the determinant expanding along the first row:

{{{Dr}}} = det{{{(matrix(3,3,  1, 1, 4, -1, 1, 6, 1, 1, 8))}}} = 1*det{{{(matrix(2,2, 1,6,1,8))}}} - 1*det{{{(matrix(2,2, -1,6,1,8))}}} + 4*det{{{(matrix(2,2, -1,1,1,1))}}} = 1*(8-6) - 1*(-8-6) + 4*(-1-1) = 2 + 14 - 8 = 8.

Hence, &nbsp;{{{p}}} = {{{Dp/D}}} = {{{4/4}}} = {{{1}}}, &nbsp;{{{q}}} = {{{Dq/D}}} = {{{20/4}}} = {{{5}}}&nbsp; and &nbsp;{{{r}}} = {{{Dr/D}}} = {{{8/4}}} = {{{2}}}.

Therefore, &nbsp;{{{x}}} = {{{1/p}}} = {{{1}}}, &nbsp;{{{y}}} = {{{1/q}}} = {{{1/5}}}&nbsp; and &nbsp;{{{z}}} = {{{1/r}}} = {{{1/2}}}.

You can easily check this solution by substituting the found values into the original system of equations. &nbsp;The solution is unique.

<B>Answer</B>. &nbsp;{{{x}}} = {{{1}}}, &nbsp;{{{y}}} = {{{1/5}}}&nbsp; and &nbsp;{{{z}}} = {{{1/2}}}.

My lessons in this site on determinants of &nbsp;3x3-matrices and the Cramer's rule for solving systems of linear equations in three unknowns are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Determinant-of-a-3x3-matrix.lesson>Determinant of a 3x3 matrix</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Co-factoring-a-3x3-determinant.lesson>Co-factoring the determinant of a 3x3 matrix</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/HOW-TO-solve-system-of-linear-eqns-in-three-unknowns-using-det.lesson>HOW TO solve system of linear equations in three unknowns using determinant (Cramer's rule)</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-word-problems-by-reducing-them-to-systems-of-linear-equations-in-three-unknowns.lesson>Solving word problems by reducing to systems of linear equations in three unknowns</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/The-trick-to-solve-some-word-problems-with-three-and-more-unknowns.lesson>The tricks to solve some word problems with three and more unknowns using mental math</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-systems-of-non-linear-equations-in-three-unknowns-using-Cramer%27s-rule.lesson>Solving systems of non-linear equations in three unknowns using Cramer's rule</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Sometime-two-eqns-are-enough-to-find-three-unknowns-by-an-UNIQUE-way.lesson>Sometime two equations are enough to find three unknowns by an UNIQUE way</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Two-very-different-approaches-to-one-word-problem.lesson>Two very different approaches to one word problem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-word-problems-in-three-unknowns-by-the-backward-method.lesson>Solving word problems in three unknowns by the backward method</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-a-system-of-linear-equations-in-17-unknowns.lesson>Solving system of linear equation in 17 unknowns</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-a-system-of-linear-equations-in-19-unknowns.lesson>Solving system of linear equation in 19 unknowns</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/OVERVIEW-of-LESSONS-on-dets-of-3x3-matrices-and-Cramer%27s-rule-for-systems-in-3-unknowns.lesson>OVERVIEW of LESSONS on determinants of 3x3-matrices and Cramer's rule for systems in 3 unknowns</A>

under the current topic &nbsp;<B>Matrices, determinant, Cramer rule</B>&nbsp; of the section &nbsp;<B>Algebra-II</B>.

My other lessons in this site on solving systems of linear equations in three unknowns are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-systems-of-linear-equations-in-3-unknowns-by-the-Substitution-method.lesson>Solving systems of linear equations in 3 unknowns by the Substitution method</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/BRIEFLY-on-solving-systems-of-linear-eqns-in-3-unknowns-by-the-Subst-method.lesson>BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/Solving-systems-of-linear-equations-in-3-unknowns-by-the-Elimination-method.lesson>Solving systems of linear equations in 3 unknowns by the Elimination method</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Matrices-and-determiminant/BRIEFLY-on-solving-systems-of-linear-eqns-in-3-unknowns-by-the-Eliminat-method.lesson>BRIEFLY on solving systems of linear equations in 3 unknowns by the Elimination method</A>

under the current topic &nbsp;<B>Matrices, determinant, Cramer rule</B>&nbsp; of the section &nbsp;<B>Algebra-II</B>.

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-II.