Answer to Question #114682 in Linear Algebra for Grant Spaans

Question

Let a11 x1 + a12 x2 + a13x3 = b1
a21 x1 + a22 x2 + a23 x3 = b2
a31 x1 + a32 x2 + a33 x3 = b3.
Show that if det (A) 6= 0 where det(A) is the determinant of the coefficient matrix;
then x2 = det(A2)
det(A) where det(A2) is the determinant obtained by replacing the second column of det(A)
by (b1; b2; b3)T :
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Answer 1

Let us rewrite the linear system in a matrix form

$\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} =\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$ or $\,\, A \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} =\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}.$

We know that $\det (A) \ne 0,$ so we may write

$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} =A^{-1}\cdot\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \dfrac{1}{\det (A)} \begin{pmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \\ \end{pmatrix} \cdot \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} .$

Here $A_{ij}$ are the elements of an adjugate matrix (https://en.wikipedia.org/wiki/Adjugate_matrix). Therefore, according to the rules of matrix multiplication we get

$x_2 = \dfrac{1}{\det (A)}\cdot (A_{12}b_1 + A_{22}b_2 + A_{32}b_3)=\\ =\dfrac{1}{\det (A)}\cdot\big(-(a_{21}a_{33}-a_{31}a_{23})b_1 + (a_{11}a_{33}-a_{31}a_{13})b_2 - (a_{11}a_{23}-a_{21}a_{13})b_3\big).$ (1)

Now let us compute determinant obtained by replacing the second column of det(A) by (b1; b2; b3)T:

$\det\begin{pmatrix} a_{11} & b_1 & a_{13} \\ a_{21} & b_2 & a_{23} \\ a_{31} & b_3 & a_{33} \\ \end{pmatrix} = a_{11}(b_2a_{33} - b_3a_{23}) - b_1(a_{21}a_{33}-a_{31}a_{23}) + a_{13}(a_{21}b_3-a_{31}b_2) = \\ = -b_1(a_{21}a_{33}-a_{31}a_{23}) +b_2(a_{11}a_{33}-a_{13}a_{31}) -b_3(a_{11}a_{23}-a_{13}a_{21}).$

If we divide the last expression by $\det (A)$ , we'll obtain the expression for $x_2,$ 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)