Solution:

$Let A=\left[\begin{array}{ccc}{1} & {1} & {1} \\ {3} & {-1} & {3} \\ {5} & {5} & {-4}\end{array}\right] \quad, \quad B=\left[\begin{array}{c}{6} \\ {10} \\ {3}\end{array}\right], \quad X=\left[\begin{array}{c}{x} \\ {y} \\ {z}\end{array}\right]$

$|A|=\left|\begin{array}{ccc}{1} & {1} & {1} \\ {3} & {-1} & {3} \\ {5} & {5} & {-4}\end{array}\right|$

$|A|=1(4-15)-1(-12-15)+1(15+5)$

$\therefore|A|=36 \neq 0$

$\therefore A^{-1} exists$

$Matrix of minors = \quad \begin{bmatrix} \quad \begin{vmatrix} -1 & 3 \\ 5 & -4 \end{vmatrix} & \quad \begin{vmatrix} 3 & 3 \\ 5 & -4 \end{vmatrix} & \quad \begin{vmatrix} 3 & -2 \\ 5 & 5 \end{vmatrix} \\ \quad \begin{vmatrix} 1 & 1 \\ 5 & -4 \end{vmatrix} & \quad \begin{vmatrix} 1 & 1 \\ 5 & -4 \end{vmatrix} & \quad \begin{vmatrix} 1 & 1 \\ 5 & 5 \end{vmatrix}\\ \quad \begin{vmatrix} 1 & 1 \\ -1 & 3 \end{vmatrix} & \quad \begin{vmatrix} 1 & 1 \\ 3 & 3 \end{vmatrix} & \quad \begin{vmatrix} 1 & 1 \\ 3 & -1 \end{vmatrix} \\ \end{bmatrix}$

$Matrix of minors =\left[\begin{array}{ccc}{-11} & {-27} & {20} \\ {-9} & {-9} & {0} \\ {4} & {0} & {-4}\end{array}\right]$

$Matrix of cofactors =\left[\begin{array}{ccc}{-11} & {27} & {20} \\ {9} & {-9} & {0} \\ {4} & {0} & {-4}\end{array}\right]$

$\begin{array}{l}{\operatorname{Adj} . A=\left[\begin{array}{ccc}{-11} & {9} & {4} \\ {27} & {-9} & {0} \\ {20} & {0} & {-4}\end{array}\right]} \\ {A^{-1}=} {\frac{1}{|A|} \text { Adj. } A}\end{array}$

$A^{-1}=\frac{1}{36}\left[\begin{array}{ccc}{-11} & {9} & {4} \\ {27} & {-9} & {0} \\ {20} & {0} & {-4}\end{array}\right]$

$\therefore X=A^{-1} B$

$\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\frac{1}{36}\left[\begin{array}{ccc}{-11} & {9} & {4} \\ {27} & {-9} & {0} \\ {20} & {0} & {-4}\end{array}\right]\left[\begin{array}{c}{6} \\ {10} \\ {3}\end{array}\right]$

$\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\frac{1}{36}\left[\begin{array}{c}{-66+90+12} \\ {162-90+0} \\ {120+0-12}\end{array}\right]$

$\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\frac{1}{36}\left[\begin{array}{c}{36} \\ {72} \\ {108}\end{array}\right]=\left[\begin{array}{l}{1} \\ {2} \\ {3}\end{array}\right]$

$\therefore \quad x=1, y=2, z=3$