Answer to Question #114474 in Linear Algebra for Mzwandile

Question

QUESTION 5

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) ≠ 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

Answer 1

Show that if det (A) ≠ 0 where det(A) is the determinant of the coefficient matrix

Let

$a_{11} x_{1} + a_{12} x_{2} + a_{13}x_{3} = b_{1}\\ a_{21} x_{1} + a_{22} x_{2} + a_{23}x_{3} = b_{2}\\ a_{31} x_{1} + a_{32} x_{2} + a_{33}x_{3} = b_{3}\\$

coefficient matrix of A is given by:

$A=\begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21} & a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix}$

$det(A)$

$det(A)= a_{11}\begin{vmatrix} a_{22}& a_{23} \\ a_{32}& a_{33} \end{vmatrix}-a_{12}\begin{vmatrix} a_{21}& a_{23} \\ a_{31} & a_{33} \end{vmatrix}+a_{13}\begin{vmatrix} a_{21}& a_{22} \\ a_{31} & a_{32} \end{vmatrix}$

$det(A)=a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{21}a_{32}-a_{31}a_{22})$

$det(A)=a_{11}a_{22}a_{33}-a_{11}a_{32}a_{23}-a_{12}a_{21}a_{33}+a_{12}a_{31}a_{23}+a_{13}a_{21}a_{32}-a_{13}a_{31}a_{22}$

hence

det(A) is not equal to zero

solving by the property of determinant:

$x_{2}det(A)=\begin{vmatrix} a_{11}&x_{2}a_{12}&a_{13} \\ a_{21} & x_{2}a_{22}&a_{23}\\ a_{31}&x_{2}a_{32}&a_{33} \end{vmatrix}$

applying the operation

$c_{2}=c_{2}+x_{1}c_{1}+x_{3}c_{3}\\x_{2}det(A)=\begin{vmatrix} a_{11}&a_{11} x_{1} + a_{12} x_{2} + a_{13}x_{3}&a_{13} \\ a_{21} & a_{21} x_{1} + a_{22} x_{2} + a_{23}x_{3}&a_{23}\\ a_{31}&a_{31} x_{1} + a_{32} x_{2} + a_{33}x_{3}&a_{33} \end{vmatrix}\\$

$\implies\\x_{2}det(A)=\begin{vmatrix} a_{11}&b_{1}&a_{13} \\ a_{21} & b_{2}&a_{23}\\ a_{31}&b_{3}&a_{33} \end{vmatrix}=det(A_{2})$

since

$x_{2}det(A)=det(A_{2}) \implies x_{2}=\frac{det(A_{2})}{det(A)}$ and det(A) is not equal to zero.