Let us rewrite the linear system in a matrix form
( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ) ⋅ ( x 1 x 2 x 3 ) = ( b 1 b 2 b 3 ) \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \\
\end{pmatrix}
\cdot \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
=\begin{pmatrix}
b_1 \\
b_2 \\
b_3
\end{pmatrix} ⎝ ⎛ a 11 a 21 a 31 a 12 a 22 a 32 a 13 a 23 a 33 ⎠ ⎞ ⋅ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ b 1 b 2 b 3 ⎠ ⎞ or A ⋅ ( x 1 x 2 x 3 ) = ( b 1 b 2 b 3 ) . \,\, A
\cdot \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
=\begin{pmatrix}
b_1 \\
b_2 \\
b_3
\end{pmatrix}. A ⋅ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ b 1 b 2 b 3 ⎠ ⎞ .
We know that det ( A ) ≠ 0 , \det (A) \ne 0, det ( A ) = 0 , so we may write
( x 1 x 2 x 3 ) = A − 1 ⋅ ( b 1 b 2 b 3 ) = 1 det ( A ) ( A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 ) ⋅ ( b 1 b 2 b 3 ) . \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}
=A^{-1}\cdot\begin{pmatrix}
b_1 \\
b_2 \\
b_3
\end{pmatrix} = \dfrac{1}{\det (A)} \begin{pmatrix}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33} \\
\end{pmatrix}
\cdot \begin{pmatrix}
b_1 \\
b_2 \\
b_3
\end{pmatrix} . ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = A − 1 ⋅ ⎝ ⎛ b 1 b 2 b 3 ⎠ ⎞ = det ( A ) 1 ⎝ ⎛ A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 ⎠ ⎞ ⋅ ⎝ ⎛ b 1 b 2 b 3 ⎠ ⎞ .
Here A i j A_{ij} A ij are the elements of an adjugate matrix (https://en.wikipedia.org/wiki/Adjugate_matrix). Therefore, according to the rules of matrix multiplication we get
x 2 = 1 det ( A ) ⋅ ( A 12 b 1 + A 22 b 2 + A 32 b 3 ) = = 1 det ( A ) ⋅ ( − ( a 21 a 33 − a 31 a 23 ) b 1 + ( a 11 a 33 − a 31 a 13 ) b 2 − ( a 11 a 23 − a 21 a 13 ) b 3 ) . x_2 = \dfrac{1}{\det (A)}\cdot (A_{12}b_1 + A_{22}b_2 + A_{32}b_3)=\\
=\dfrac{1}{\det (A)}\cdot\big(-(a_{21}a_{33}-a_{31}a_{23})b_1 + (a_{11}a_{33}-a_{31}a_{13})b_2 - (a_{11}a_{23}-a_{21}a_{13})b_3\big). x 2 = det ( A ) 1 ⋅ ( A 12 b 1 + A 22 b 2 + A 32 b 3 ) = = det ( A ) 1 ⋅ ( − ( a 21 a 33 − a 31 a 23 ) b 1 + ( a 11 a 33 − a 31 a 13 ) b 2 − ( a 11 a 23 − a 21 a 13 ) b 3 ) . (1)
Now let us compute determinant obtained by replacing the second column of det(A) by (b1; b2; b3)T:
det ( a 11 b 1 a 13 a 21 b 2 a 23 a 31 b 3 a 33 ) = a 11 ( b 2 a 33 − b 3 a 23 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) = = − b 1 ( a 21 a 33 − a 31 a 23 ) + b 2 ( a 11 a 33 − a 13 a 31 ) − b 3 ( a 11 a 23 − a 13 a 21 ) . \det\begin{pmatrix}
a_{11} & b_1 & a_{13} \\
a_{21} & b_2 & a_{23} \\
a_{31} & b_3 & a_{33} \\
\end{pmatrix}
= a_{11}(b_2a_{33} - b_3a_{23}) - b_1(a_{21}a_{33}-a_{31}a_{23}) + a_{13}(a_{21}b_3-a_{31}b_2) = \\
= -b_1(a_{21}a_{33}-a_{31}a_{23}) +b_2(a_{11}a_{33}-a_{13}a_{31}) -b_3(a_{11}a_{23}-a_{13}a_{21}). det ⎝ ⎛ a 11 a 21 a 31 b 1 b 2 b 3 a 13 a 23 a 33 ⎠ ⎞ = a 11 ( b 2 a 33 − b 3 a 23 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) = = − b 1 ( a 21 a 33 − a 31 a 23 ) + b 2 ( a 11 a 33 − a 13 a 31 ) − b 3 ( a 11 a 23 − a 13 a 21 ) .
If we divide the last expression by det ( A ) \det (A) det ( A ) , we'll obtain the expression for x 2 , x_2, x 2 , that we've obtained above (see (1)).
Note that we've obtained a formula from Cramer's rule (see https://en.wikipedia.org/wiki/Cramer's_rule#General_case)