Question

solve the following system of linear equations using Cramer's rule x+y=0, y+z=1, z+x=3

Answer 1

take, x+y+y+z+z+x=4
(x+y)+(x+y)+2z=4
0+0+2z=4
z=2
z+x=3
x=3-2=1
x+y=0
1+y=0
y=-1
I think this may help you

Answer 2

→The given system of equations are:

x+y =0

y+z = 1

z+x=3

The system of equation written in Matrix form is given by

→A X = B, where A = $\begin{pmatrix}1 &1  &0 \\ 0 &1  &1 \\ 1 &0  &1 \end{pmatrix}\end{matrix}$, X =$\begin{pmatrix}x\\ y\\ z\end{pmatrix}$ and B = $\begin{pmatrix}0\\ 1\\ 3\end{pmatrix}$

→X = $A^{-1}B$

and , $A^{-1}=\frac{Adjoint A}{Determinant A}$

Determinant A = 1 ×(1-0) -1(0-1)=1+1=2

To find Adjoint A, we will find the cofactor of matrix A and then take transpose of it.

→Cofactor of Matrix A = Transpose of $\begin{pmatrix}1 &1  &-1\\ -1 &1  &1 \\ 1 &-1  &1 \end{pmatrix}\end{matrix}$ =$\begin{pmatrix}1 &-1  &1 \\ 1 &1  &-1 \\ -1 &1  &1 \end{pmatrix}\end{matrix}$

→Now, $A^{-1}B =\begin{pmatrix}2\\ -2\\ 4\end{pmatrix}$

→$\frac{A^{-1}.B}{Det.A}=\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\frac{\begin{pmatrix}2\\ -2\\ 4\end{pmatrix}}{2}\\\\<strong>x=1, y=-1,z=2,$ which is desired result.