# How do you solve the equations below using Cramer's rule? 2x+y-3z=3 3x-2y+4z=2 4x+2y-6z=-7

lemjay | High School Teacher | (Level 3) Senior Educator

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The equations are:

(I) `2x + y -3z = 3`

(II) `3x-2y+4z=2`

(III) `4x+2y-6z=-7`

Using Cramer's Rule, first we need to solve the determinant of the coefficient matrix, D.

Note that the size of the matrix is 3x3 since there are three unknown variables. It consists of coefficients of x, y and z written under the first, second and third column respectively.

So,

2     1    -3

D=   3    -2     4

4      2    -6

Then, solve the detetminant of D.

2      1     -3     2      1

detD=     3     -2      4     3     -2

4       2     -6     4      2

`detD= [2*(-2)*(-6)+1*4*4+(-3)*3*2] - [4*(-2)(-3)+2*4*2+(-6)*3*1] `

`detD = 0`

Next, we set-up the matrix Dx. This can be done by replacing the coefficients of x with the constants of the given equations.

3      1     -3

Dx=    2    -2      4

-7      2     -6

Then, solve the determinant of Dx.

3      1     -3       3       1

Dx=    2     -2       4       2     -2

-7      2      -6      -7      2

`detD_x=[3(-2)(-6)+1*4(-7)+(-3)*2*2]-[(-7)(-2)(-3)+2*4*3+(-6)*2*1]`

`detD_x=26`

Set-up matrix Dy too. So replace the coefficients of y with the constants.

2    3    -3

Dy=  3     2     4

4    -7    -6

Solve for detDy.

2    3    -3     2      3

Dy=  3     2     4     3      2

4    -7    -6     4    -7

`detD_y= [2*2(-6)+3*4*4+(-3)3(-7)]-[4*2(-3)+(-7)*4*2+(-6)*3*3]`

`detD_y=221`

Also, set-up matrix Dz. So replace coefficients of z with constants.

2      1      3

Dz=   3     -2      2

4      2     -7

Solve for Dz.

2      1      3      2     1

Dz=   3     -2      2      3    -2

4      2     -7      4     2

`detD_z=[2(-2)(-7)+1*2*4+3*3*2]-[4(-2)*3+2*2*2+(-7)*3*1]`

`detD_z=91`

Note that if detD is zero and the determinants of Dx, Dy and Dz are not equal to zero, then the system of equations has no solution.

Hence,  2x+y-3z=3, 3x-2y+4z=2 and 4x+2y-6z=-7 has no solution.

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