# Thread: Solve this 4x4 matrix using Cramer's rule

1. ## Solve this 4x4 matrix using Cramer's rule

(1-a1-a2 ; a1 ; a2 ; 0)
(b1 ; 1-b1-a2+c2 ; 0 ; a2-c2)
(b2 ; 0 ; 1-b2-a1+c2 ; a1-c1)
(0 ; b2+c2 ; b1+c1 ; 1-b1-c1-b2-c2)

all 4 rows equal 1 and the variables to be found (going across the rows) are w, x, y and z.

When I tried to solve the first determinant I got 4 brackets with 3, 4, 4 and 5 elements in and the multiplying out got too confusing so I am not sure if it was right or not.

2. Originally Posted by Sarahstar
(1-a1-a2 ; a1 ; a2 ; 0)
(b1 ; 1-b1-a2+c2 ; 0 ; a2-c2)
(b2 ; 0 ; 1-b2-a1+c2 ; a1-c1)
(0 ; b2+c2 ; b1+c1 ; 1-b1-c1-b2-c2)

all 4 rows equal zero and the variables to be found (going across the rows) are w, x, y and z.

When I tried to solve the first determinant I got 4 brackets with 3, 4, 4 and 5 elements in and the multiplying out got too confusing so I am not sure if it was right or not.
I'm going to see if I can clean this matrix up a bit.

$\begin{pmatrix} 1 - a_1 - a_2 & a_1 & a_2 & 0 \\ b_1 & 1 - b_1 - a_2 + c_2 & 0 & a_2 - c_2 \\ b_2 & 0 & 1 - b_2 - a_1 + c_2 & a_1 - c_1 \\ 0 & b_2 + c_2 & b_1 + c_1 & 1 - b_1 - c_1 - b_2 - c_2 \end{pmatrix} \begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$

Is this what you mean?

-Dan

3. yes that's correct. sorry I didn't know how to write a matrix on here

4. actually, sorry I just realise the matrix is on it's side so the columns should be the rows and vice versa:
(1-a1-a2 ; b1 ; b2 ; 0)
(a1 ; 1-b1-a2+c2 ; 0 ; b2+c2)
etc