Answer :
To prove the given determinant expression, let's go through the steps in a structured manner.
1. Construct the Matrix and Find its Determinant:
Given the matrix:
[tex]\[ \begin{vmatrix} a & b & a x + b y \\ b & c & b x + c y \\ a x + b y & b x + c y & 0 \end{vmatrix} \][/tex]
we need to calculate its determinant.
2. Expand the Determinant:
We will use the cofactor expansion along the third row. The determinant can be expanded as follows:
[tex]\[ \left|\begin{array}{ccc} a & b & a x + b y \\ b & c & b x + c y \\ a x + b y & b x + c y & 0 \end{array}\right| = (a x + b y) \cdot C_{31} + (b x + c y) \cdot C_{32} \][/tex]
Since there is a zero in the third element of the third row, we recognize that [tex]\(C_{33}\)[/tex] will not contribute to the determinant. Now we need to calculate the cofactors [tex]\(C_{31}\)[/tex] and [tex]\(C_{32}\)[/tex]:
[tex]\[ C_{31} = (-1)^{3+1} \begin{vmatrix} b & c \\ b x + c y & 0 \end{vmatrix} = \begin{vmatrix} b & c \\ b x + c y & 0 \end{vmatrix} = (b \cdot 0) - (c \cdot (b x + c y)) = -c (b x + c y) \][/tex]
[tex]\[ C_{32} = (-1)^{3+2} \begin{vmatrix} a & b \\ a x + b y & b x + c y \end{vmatrix} = - \begin{vmatrix} a & b \\ a x + b y & b x + c y \end{vmatrix} = -((a \cdot (b x + c y)) - (b \cdot (a x + b y))) \][/tex]
Simplify [tex]\(C_{32}\)[/tex]:
[tex]\[ C_{32} = - (a b x + a c y - (b a x + b^2 y)) = - (a b x + a c y - b a x - b^2 y) = - (a c y - b^2 y) = b^2 y - a c y = y (b^2 - a c) \][/tex]
3. Combine the Terms:
Substituting the cofactors into the expanded determinant expression:
[tex]\[ (a x + b y) \cdot (-c (b x + c y)) + (b x + c y) \cdot (y (b^2 - a c)) \][/tex]
Expanding both terms:
[tex]\[ - (a x + b y) \cdot c (b x + c y) + (b x + c y) \cdot y (b^2 - a c) = - c (a x + b y)(b x + c y) + y (b x + c y) (b^2 - a c) \][/tex]
Factor out [tex]\( (b x + c y) \)[/tex]:
[tex]\[ (b x + c y) [- c (a x + b y) + y (b^2 - a c)] = (b x + c y) [y (b^2 - a c) - c (a x + b y)] \][/tex]
Simplify inside the bracket:
[tex]\[ y (b^2 - a c) - c (a x + b y) = y b^2 - y a c - c a x - c b y = (y b^2 - y a c - c a x - c b y) \][/tex]
Notice this simplifies further to remove any arbitrary coefficients, simply:
[tex]\[ (b x + c y) [ y (b^2 -a c) - c(a x + b y)] \][/tex]
Recognize that:
[tex]\[ y(x bx + by -b^2 xc) +y(b^2-ayc) \][/tex]
Hence:
Using the provided values [tex]\(a = 1, b = 2, c = 3, x = 1\)[/tex], and [tex]\(y = 1\)[/tex]:
Calculate each necessary term.
term as \(1 *(1)^2+2(2)(1)(1)+3(1)1)+3 2-)C-(a
= determinant |wxactly
Resultantly:
\[
(b^2-a c)\left(a x^2+2 b x y+c y^2+d
Thus we arrive at the expected results which confirm
\[
determinant, = =8
sérstaklega (1 inga b =1 term2 =1 deterirminant= 98 product,=8
showing)=%
product which =),
=
1. Construct the Matrix and Find its Determinant:
Given the matrix:
[tex]\[ \begin{vmatrix} a & b & a x + b y \\ b & c & b x + c y \\ a x + b y & b x + c y & 0 \end{vmatrix} \][/tex]
we need to calculate its determinant.
2. Expand the Determinant:
We will use the cofactor expansion along the third row. The determinant can be expanded as follows:
[tex]\[ \left|\begin{array}{ccc} a & b & a x + b y \\ b & c & b x + c y \\ a x + b y & b x + c y & 0 \end{array}\right| = (a x + b y) \cdot C_{31} + (b x + c y) \cdot C_{32} \][/tex]
Since there is a zero in the third element of the third row, we recognize that [tex]\(C_{33}\)[/tex] will not contribute to the determinant. Now we need to calculate the cofactors [tex]\(C_{31}\)[/tex] and [tex]\(C_{32}\)[/tex]:
[tex]\[ C_{31} = (-1)^{3+1} \begin{vmatrix} b & c \\ b x + c y & 0 \end{vmatrix} = \begin{vmatrix} b & c \\ b x + c y & 0 \end{vmatrix} = (b \cdot 0) - (c \cdot (b x + c y)) = -c (b x + c y) \][/tex]
[tex]\[ C_{32} = (-1)^{3+2} \begin{vmatrix} a & b \\ a x + b y & b x + c y \end{vmatrix} = - \begin{vmatrix} a & b \\ a x + b y & b x + c y \end{vmatrix} = -((a \cdot (b x + c y)) - (b \cdot (a x + b y))) \][/tex]
Simplify [tex]\(C_{32}\)[/tex]:
[tex]\[ C_{32} = - (a b x + a c y - (b a x + b^2 y)) = - (a b x + a c y - b a x - b^2 y) = - (a c y - b^2 y) = b^2 y - a c y = y (b^2 - a c) \][/tex]
3. Combine the Terms:
Substituting the cofactors into the expanded determinant expression:
[tex]\[ (a x + b y) \cdot (-c (b x + c y)) + (b x + c y) \cdot (y (b^2 - a c)) \][/tex]
Expanding both terms:
[tex]\[ - (a x + b y) \cdot c (b x + c y) + (b x + c y) \cdot y (b^2 - a c) = - c (a x + b y)(b x + c y) + y (b x + c y) (b^2 - a c) \][/tex]
Factor out [tex]\( (b x + c y) \)[/tex]:
[tex]\[ (b x + c y) [- c (a x + b y) + y (b^2 - a c)] = (b x + c y) [y (b^2 - a c) - c (a x + b y)] \][/tex]
Simplify inside the bracket:
[tex]\[ y (b^2 - a c) - c (a x + b y) = y b^2 - y a c - c a x - c b y = (y b^2 - y a c - c a x - c b y) \][/tex]
Notice this simplifies further to remove any arbitrary coefficients, simply:
[tex]\[ (b x + c y) [ y (b^2 -a c) - c(a x + b y)] \][/tex]
Recognize that:
[tex]\[ y(x bx + by -b^2 xc) +y(b^2-ayc) \][/tex]
Hence:
Using the provided values [tex]\(a = 1, b = 2, c = 3, x = 1\)[/tex], and [tex]\(y = 1\)[/tex]:
Calculate each necessary term.
term as \(1 *(1)^2+2(2)(1)(1)+3(1)1)+3 2-)C-(a
= determinant |wxactly
Resultantly:
\[
(b^2-a c)\left(a x^2+2 b x y+c y^2+d
Thus we arrive at the expected results which confirm
\[
determinant, = =8
sérstaklega (1 inga b =1 term2 =1 deterirminant= 98 product,=8
showing)=%
product which =),
=