We are given,

ax + by = c   ....(1)

bx + ay = 1 + c   ....(2)

Now, we can generate the matrix.

$\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}\end{array}\right]\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{c}\mathrm{c}\\ 1+\mathrm{c}\end{array}\right]$

Let $\mathrm{A}=\left[\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\\ \mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}\end{array}\right]$

So, $\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]={\mathrm{A}}^{-1}\left[\begin{array}{c}\mathrm{c}\\ 1+\mathrm{c}\end{array}\right]$

Now, ${\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\left[\begin{array}{cc}\mathrm{a}& -\mathrm{b}\\ -\mathrm{b}& \mathrm{a}\end{array}\right]$

here |A| = (a) (a) – (b) (b) = a² – b²

So, $\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\frac{1}{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}\left[\begin{array}{cc}\mathrm{a}& -\mathrm{b}\\ -\mathrm{b}& \mathrm{a}\end{array}\right]\left[\begin{array}{c}\mathrm{c}\\ 1+\mathrm{c}\end{array}\right]$

$\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\frac{1}{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}\left[\begin{array}{c}\mathrm{ac}-\mathrm{b}-\mathrm{bc}\\ -\mathrm{bc}+\mathrm{a}+\mathrm{ac}\end{array}\right]$

So, the value of x is $\frac{\mathrm{ac}-\mathrm{b}-\mathrm{bc}}{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}$ &

the value of y is $\frac{\mathrm{a}+\mathrm{ac}-\mathrm{bc}}{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}$