Answer to Question #110825 in Linear Algebra for Hetisani Sewela

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

Answer 1

Show that if det (A) 6= 0 where det(A) is the determinant of the coefficient matrix

coefficient matrix of the system of equations is

a11 x1 + a12 x2 + a13x3 = b1

a21 x1 + a22 x2 + a23 x3 = b2

a31 x1 + a32 x2 + a33 x3 = b3

then the coefficient matrix A is

A= $\begin{bmatrix} a11& a12&a13\\ a21& a22&a23\\ a31&a32&a33 \end{bmatrix}$

det(A)6=[a₁₁(a₂₂a₃₃-a₃₂a₂₃)-a₁₂(a₂₁a₃₃-a₃₂a₂₃)₊a₁₃(a₂₁a₃₂-a₃₁a₂₂)]6

₌[[a₁₁(a₂₂a₃₃-a₃₂a₂₃)_-a₁₂(a₂₁a₃₃-a₃₂a₂₃)₊a₁₃(a₂₁a₃₂-a₃₁a₂₂]

_{let Y=}(a₂₂a₃₃-a₃₂a₂₃) ; Z=a₂₁a₃₃-a₃₂a_{23 ; L=}a₂₁a₃₂-a₃₁a₂₂

det(A)6=[a₁₁Y-a₁₂Z₊a₁₃L]6

det(A)^6=[a₁₁Y-a₁₂Z₊a₁₃L]6 which is not equal to 0

for: x2=det(A2)

Since we need to replace the second column by (b1,b2,b3), so we first multiply the second column by x2 (when we multiply any determinant by a scalar, it either gets multiplied with a single row or with a single column). Then used the operation to obtain (b1,b2,b3) in second row which you u can easily observe in my solution

then

from the system of linear equations below

a11 x1 + a12 x2 + a13x3 = b1

a21 x1 + a22 x2 + a23 x3 = b2

a31 x1 + a32 x2 + a33 x3 = b3

then:

$x2det(A)=\begin{vmatrix} a11 & x2a12&a13 \\ a21 & x2a22&a23\\ a31&x2a32&a33 \end{vmatrix}$

taking the operation C₂¹=C₂+x1C₁+x3C₃

we have

$x2det(A)=\begin{vmatrix} a11 & a11 x1 + a12 x2 + a13x3&a13 \\ a21 & a21 x1 + a22 x2 + a23 x3&a23\\ a31&a31 x1 + a32 x2 + a33 x3&a33 \end{vmatrix}$

$x2det(A)=\begin{vmatrix} a11 & b1&a13 \\ a21 & b2&a23\\ a31&b3&a33 \end{vmatrix}=det(A2)$

hence

$x2det(A)=det(A2)\\ then\\ x2=det(A2)/det(A)\\$ where det(A) is not equal to zero

if det(A)=1 then x2=det(A2)