Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 3 / Solution

To solve $A\mathbf{x}=\mathbf{b}$, subtract the first row from the second, and twice the first from the third. The resulting equations give $y=b_2-b_1$, $y+z=b_3-2b_1$, and hence $z=b_3-b_1-b_2$. Substituting in the original first equation gives $x=7b_1-3b_3$. Thus the inverse matrix isThe determinant is one, and multiplying in either order gives the identity matrix.