Answer to Question #114479 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

Let $a_i=(a_{1i}\,\,\,\, a_{2i}\,\,\,\,a_{3i})^T$ for $i=1,2,3$, and $b=(b_1\,\,\,\,b_2\,\,\,\,b_3)^T$, $x=(x_1\,\,\,\,x_2\,\,\,\,x_3)^T$.

We have $b=Ax=a_1x_1+a_2x_2+a_3x_3$.

Then $|a_1 \,\,\,\, b\,\,\,\,a_3|=|a_1\,\,\,\, (a_1x_1+a_2x_2+a_3x_3) \,\,\,\,a_3|=$

$=x_1|a_1\,\,\,\,a_1\,\,\,\,a_3|+x_2|a_1\,\,\,\,a_2\,\,\,\,a_3|+x_3|a_1\,\,\,\,a_3\,\,\,\,a_3|=$

$=x_2 \det A$.

If $\det A \ne 0$,  it follows that $x_2=\frac{|a_1 \,\,\,\,b\,\,\,\,a_3|}{\det A}$ .