- #1

lendav_rott

- 232

- 10

## Homework Statement

2x +5y -z = 3

3x + 5y -4z = -6 the bracket needs to go infront, so x,y,z satisfy all the equations at once.

4x -2y +6z = 1

## Homework Equations

Cramer's method:

(I don't know how to add certain symbols to the text so it is difficult to express it)

If A were the matrix of the coefficients of the variables then a

_{ij}is a corresponding element and A

_{ij}= (-1)

^{i+j}* M

_{ij}- the subdeterminant? Is it the correct word? M

_{ij}is a minor of the corresponding element.

So Cramer's idea is this:

|A| = a

_{11}* A

_{11}+ a

_{21}* A

_{21}+ a

_{31}* A

_{31}

and

a

_{12}* A

_{11}+ a

_{22}* A

_{21}+ a

_{32 }* A

_{31}= 0

and

a

_{13}* A

_{11}+ a

_{23}* A

_{21}+ a

_{33}* A

_{31}= 0

## The Attempt at a Solution

I have no problem to solve this system, but the problem is:

I get the correct determinant value , but the following equations (which need to be 0) are not zero. Is Cramer's idea flawed? I am certain I have done the calculations correctly.

|A| = 2 * 38 - 3*28 - 4*15 = -68 good

5*38 - 5*28 + 2*15 =/= 0 not good

-38 + 4*28 - 6*15 =/= 0 definitely not good

I can also skip to the part where I replace the x,y,z coefficients with the numbers on the right side of the equal sign when I solve for x,y,z , but I am not so much interested in the solution, am I even allowed to do that operation when the requirements are not met?

I can also say for certain that there is a solution to it, because |A| =/= 0.

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