Answer:First we make the system as diagonally dominant, i.e., (|a_{ii}|>\sum_{j=1, i \neq j}^{3} | a_{ij}| ).

$43x+2y+25z=23$

$3x+53y+3z=91$

$2x-4y+49z=49$

$x^{(n+1)}=\dfrac{23-2y^n-25z^n}{43}$

$y^{(n+1)}=\dfrac{91-3x^{(n+1)}-3z^n}{53}$

$z^{(n+1)}=\dfrac{49-2x^{(n+1)}+4y^{(n+1)}}{49}$

Taking initial values as below:

$x^{(0)}=0,y^{(0)}=0,z^{(0)}=0$

| $n$ | $x^{(n+1)}$ | $y^{(n+1)}$ | $z^{(n+1)}$ | | --- | --- | --- | --- | | 0 1 2 3 | 0 0.5349 -0.1923 -0.2075 -0.2078 | 0 1.6867 1.6647 1.6640 1.6640 | 0 1.1159 1.1437 1.1443 1.1443 | Answer: $x=-0.208, y=1.664, z=1.144$ upto 3 decimal places.