Question #114682
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 :
[
1
Expert's answer
2020-05-08T18:53:05-0400

Let us rewrite the linear system in a matrix form

(a11a12a13a21a22a23a31a32a33)(x1x2x3)=(b1b2b3)\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(x1x2x3)=(b1b2b3).\,\, 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)0,\det (A) \ne 0, so we may write

(x1x2x3)=A1(b1b2b3)=1det(A)(A11A21A31A12A22A32A13A23A33)(b1b2b3).\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 AijA_{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

x2=1det(A)(A12b1+A22b2+A32b3)==1det(A)((a21a33a31a23)b1+(a11a33a31a13)b2(a11a23a21a13)b3).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(a11b1a13a21b2a23a31b3a33)=a11(b2a33b3a23)b1(a21a33a31a23)+a13(a21b3a31b2)==b1(a21a33a31a23)+b2(a11a33a13a31)b3(a11a23a13a21).\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)\det (A) , we'll obtain the expression for x2,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)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS