# Cramer's Rule - 3X3 Linear System

In this tutorial we're going to use Crater's rule to solve a system of linear equations with three variables. So let's say, we have 2x plus y, minus Z and that's equal to 1 and also 3 x + 2, y plus 2z and that's going to be equal to 13 and then 4x minus 2y, plus 3z, let's say, that's equal to 9. Now. This equation is in this form. So it's, a 1 X, plus B 1, y, plus C 1, z and that's equal to D 1. And then we have a 2 X, plus B 2, y plus C 2, Z is equal to D 2.

And then a 3 X, plus B, 3, y, plus C, 3 Z is equal to D 3. So the. Letters, ABC and D represents the coefficients in this equation.

And the subscripts 1 2 3 tells us the role that we're dealing with. So subscript 1 is for the first row subscript 3 is for the third row. Now X is equal to DX divided by D. Y is Dy over D and Z is DZ divided by D. So we need to calculate X, I mean, DX, Dy, DZ and D. And then we can get the solution to these equations. So let's start with D is going to be a 3 by 3 determinant. And the elements in this matrix is going to be a 1 a 2 a 3. So those. Are the coefficients for the X variables and then B 1 B 2, B 3.

So those are the coefficients for the Y variables. And then C 1 C 2, C 3. So you need to know how to evaluate 3 by 3 determinant.

So first let's start with a 1. If we eliminate Row 1 and column 1 we'll, be left with this. So it's going to be a 1 multiplied by a 2 by 2 matrix, or the determinant of a 2 by 2 matrix, rather B 2, B 3, C 2, C 3. And then it's going to be minus the next one, which is B 1 and needs to eliminate Row 1 column 2 because that's. Where it B 1 is located in Row 1 column 2, and so we're going to have a 2 a 3 and C 2 C 3 left over, and then it's going to be plus C 1 and C 1 is located in row 1 column, 3, so we're going to get this a 2, a 3 B 2, B 3 so that's how you can evaluate a 3 by 3 determinant. So in this example, a 1 is 2 and then B 2 matches with 2.

So B 2 is 2. B 3 is negative 2. C 2 is 2 and C 3 is stream. And then minus B 1 B 1 is 1 it's, the coefficient in front of Y. And then a 2 is 3. A 3 is 4.

C 2 is 2 and C 3 is 3. And then plus C 1. Which is negative 1. So C 1 is in front of Z, that's negative 1. And then a 2 that's 3. A 3 is 4. B 2 is 2.

B 3 is negative 2. So now let's find the determinant of a 2 by 2 matrix. So it's going to be 2 times 3 and then minus 2 times negative 2. And we still have a 2 in front of that.

This is going to be 3 times 3, which is 9 minus 2 times 4, which is 8. And this is going to be 3 times negative 2, which is negative 6 minus 2 times 4, which is 8 now 2 times 3 at 6, negative 2 times negative 2 is positive, 4 and 9 minus 8. Is 1 negative 6, minus 8 is negative. 14 6, plus 4 is 10 times 2 is 20 and negative, 14 that's positive 14. So we have 20 plus 14, which is 34 minus 1 that's, 33 so D is equal to 33 now let's calculate DX.

So for the elements and deal's, replace the coefficients in front of X with D 1, D 2, D 3. So it's going to be D 1, D 2, D 3 and then B 1 B 2, B 3 and C 1, C 2, C 3. So D 1 is 1. D 2 is 13 and that's 9.

And then for the B's. Those are the coefficients in front of Y. And then C 1 C 2, C 3. Those are the.

Coefficients for Z. So we're gonna start with this number it's going to be 1. And if we take out the row and column that corresponds to that number we're going to get these numbers. So that's 2 2, negative 2 and 3, and then it's going to be minus the second one, which is also 1. And then that element corresponds to Row 1 column 2, leaving behind these four elements. So it's going to be 13, 9 2 3 and then Plus this one, which is negative 1. And that corresponds to Row 1 column, 3 which leaves behind 13 to 9 and.

Negative 2 so this is going to be 1 times 2 minus 3 is 6 and then minus 2 times negative 2, which is negative 4. And then it's going to be 13 times 3, which is 39 minus 2 times 9, which is 18 and then 13 times negative 2, that's negative, 26 minus 2 times 9, which is 18 so 6 minus negative 4 and 6, plus 4, which is 10 39 minus 18 is 21 with a negative 1 in front of it. So that's negative, 21 negative, 26 minus 18 that's going to be negative 44 times negative 1 so that's, positive 44. So we have 10 plus 44, which is 54. Minus 21, which is 33, so DX is equal to 33 now let's focus on D wire. So this time, the coefficients in front of why we're going to replace it with d1 d2 and d3. So it's going to be a 1 a 2, a 3 and then D 1 D 2, D 3 and then C 1, C 2, C 3.

So this corresponds to the numbers 2, 3, 4 1, 13, 9 and negative 1 2, & 3. So let's start with this number 2. So that's in Row 1 column 1. So it leaves behind 13 - 9 3. And then minus the next number, which is 1 and that's located in Row 1 column 2.

So it leaves behind 3 2, 4 3. And then. It's going to be plus negative, one and that's in row 1 column, 3 so that's going to be 3 13, 4 9.

So this is going to be 2. And then 13 times 3 is 39 minus 2 times 9, which is 18. And then 3 + 3 is 9 2. + 4 is 8 3 times.

9 is 27 13 times 4 is 52 now 39 minus 18 is 21. 9 minus 8 is 1 27, minus 52 that's negative. 25 2 times 21 is 42 and 42. Minus 1 is 41. 41 plus 25 is 66. So Dy is 66 now let's move on to the last one DZ. So in this case, the coefficients of Z will be replaced with D 1, D 2 and D 3.

So it's going to be a 1. A 2 a 3 B 1, B 2, B 3 and then D 1, D 2, D 3. So we're going to have the numbers 2 3 4 1 2, negative 2, & 1, 13, 9 so let's start with this one too. So if we eliminate Row 1 column 1 we're going to have to 13 negative 2 9 and then minus the next number, which is 1. And then if we eliminate Row 1 column 2, we'll have 3 13, 4 9, and then it's going to be +1.

So let's eliminate Row 1 column, 3 and that's going to be 3 - 4 negative 2. So this is going to equal 2 and then 2 times 9 is 18. And then the minus negative 2 times 13. That's a negative 26. And then minus 1 3 times 9 is 27 13 times 4 is 52. And then 3 times negative 2 is negative. 6 2 times 4 is 8 now.

18 plus 26, that's, 44 27, minus 52 is negative 25 and 6 times I mean, negative 6, minus 8 is negative. 14 two times 44 is 88. Negative 1 times negative, 25 positive, 25. Now, 25 minus 14 is 11 and 88. Plus 11 is 99. So DZ is 99. Now we have everything that we need to finish the problem.

So X is equal to DX over D and that's going to be 33 divided by 33, which is 1. So that means that. X is equal to 1. Y is Dy over D and that's going to be 66 divided by 33, which is 2 and Y is equal to 2. And then Z is going to be DZ over D. So that's, 99, divided by 33, which is 3 so Z is equal to 3. So the solution is 1. Comma 2, comma 3 in the form of X, comma, Y, comma, Z and that's it.

So now you know how to use Crater's rule to solve a system of equations with three variables.

Dated : 18-Apr-2022