(a) Prove that. [3] vmatrix a & b & ax + by \\ b & c & bx + cy \\ ax + by & bx + cy & 0 vmatrix = (b^2 ac) (ax^2 + 2bxy + cy^2) (b) Ramesh sells 3 shares of A and buys 2 shares of B, thus increasing…

(a) Proof of the Determinant Identity

We are given the determinant:

$$\left|\begin{array}{ccc}a& b& ax+by\\ b& c& bx+cy\\ ax+by& bx+cy& 0\end{array}\right|$$

Let’s denote this determinant as $D$. We will expand it using the cofactor expansion along the third row (since it contains a zero, simplifying calculations).

Step 1: Cofactor Expansion

The determinant $D$ can be expanded as:

$$D=(ax+by)\cdot \left|\begin{array}{cc}b& bx+cy\\ c& 0\end{array}\right|-(bx+cy)\cdot \left|\begin{array}{cc}a& ax+by\\ b& bx+cy\end{array}\right|+0\cdot \left|\begin{array}{cc}a& b\\ b& c\end{array}\right|$$

Simplifying the minors:

1. First minor: $$\left|\begin{array}{cc}b& bx+cy\\ c& 0\end{array}\right|=b\cdot 0-(bx+cy)\cdot c=-c(bx+cy)$$

2. Second minor: $$\left|\begin{array}{cc}a& ax+by\\ b& bx+cy\end{array}\right|=a(bx+cy)-b(ax+by)=abx+acy-abx-b^{2}y=(ac-b^2)y$$

Substituting back:

$$D=(ax+by)(-c)(bx+cy)-(bx+cy)(ac-b^2)y$$

Step 2: Factor and Simplify

Factor out $(bx+cy)$:

$$D=(bx+cy)\left[-c(ax+by)-(ac-b^2)y\right]$$

Expand inside the brackets:

$$=(bx+cy)\left[-acx-bcy-acy+b^{2}y\right]$$

Factor $y$ from the last two terms:

$$=(bx+cy)\left[-acx-bcy+y(b^2-ac)\right]$$

Now, distribute $(bx+cy)$:

$$=-acx(bx+cy)-bcy(bx+cy)+y(b^2-ac)(bx+cy)$$

Simplify each term:

1. $-acx(bx+cy)=-acb{x}^2-a{c}^{2}xy$
2. $-bcy(bx+cy)=-b^{2}cxy-b{c}^2{y}^2$
3. $y(b^2-ac)(bx+cy)=(b^2-ac)(bxy+c{y}^2)=b^{3}xy+b^{2}c{y}^2-abxy-a{c}^2{y}^2$

Combine all terms:

$$D=-acb{x}^2-a{c}^{2}xy-b^{2}cxy-b{c}^2{y}^2+b^{3}xy+b^{2}c{y}^2-abxy-a{c}^2{y}^2$$

Simplify by combining like terms:

• $x^2$: $-acb{x}^2$
• $xy$: $-a{c}^{2}xy-b^{2}cxy+b^{3}xy-abxy=(-a{c}^2-b^{2}c+b^3-ab)xy$
• $y^2$: $-b{c}^2{y}^2+b^{2}c{y}^2-a{c}^2{y}^2=(-b{c}^2+b^{2}c-a{c}^2)y^2$

However, this approach seems overly complex. Instead, let’s factor the original expression differently:

Notice that the determinant can be rewritten by grouping terms. Let’s define:

$$D=\left|\begin{array}{ccc}a& b& ax+by\\ b& c& bx+cy\\ ax+by& bx+cy& 0\end{array}\right|$$

Let $u=ax+by$ and $v=bx+cy$. Then:

$$D=\left|\begin{array}{ccc}a& b& u\\ b& c& v\\ u& v& 0\end{array}\right|$$

Expanding along the third row:

$$D=u\left|\begin{array}{cc}b& v\\ c& 0\end{array}\right|-v\left|\begin{array}{cc}a& u\\ b& v\end{array}\right|+0\left|\begin{array}{cc}a& b\\ b& c\end{array}\right|$$

$$=u(-cv)-v(av-bu)=-ucv-a{v}^2+bvu$$

Substitute back $u$ and $v$:

$$=-(ax+by)(bx+cy)c-a(bx+cy{)}^2+b(ax+by)(bx+cy)$$

Factor out $(bx+cy)$:

$$=(bx+cy)\left[-c(ax+by)-a(bx+cy)+b(ax+by)\right]$$

Simplify inside the brackets:

$$=(bx+cy)\left[-acx-bcy-abx-acy+abx+b^{2}y\right]$$

$$=(bx+cy)\left[-acx-bcy-acy+b^{2}y\right]$$

Factor $y$ from the last two terms:

$$=(bx+cy)\left[-acx+y(b^2-ac-bc)\right]$$

This still seems messy. Instead, let’s use the property of determinants for matrices of the form:

$$\left|\begin{array}{ccc}a& b& u\\ b& c& v\\ u& v& 0\end{array}\right|$$

This is a skew-symmetric-like determinant, and its value can be shown to be:

$$D=(b^2-ac)(u^2+v^2)-(bu-av{)}^2$$

But this seems incorrect. Instead, let’s recompute carefully:

From the expansion:

$$D=-ucv-a{v}^2+bvu$$

Substitute $u=ax+by$ and $v=bx+cy$:

$$D=-(ax+by)(bx+cy)c-a(bx+cy{)}^2+b(ax+by)(bx+cy)$$

Let’s expand $(ax+by)(bx+cy)$:

$$=ab{x}^2+acxy+b^{2}xy+bc{y}^2=ab{x}^2+(ac+b^2)xy+bc{y}^2$$

Now substitute back:

$$D=-c\left[ab{x}^2+(ac+b^2)xy+bc{y}^2\right]-a\left[b^2{x}^2+2bcxy+c^2{y}^2\right]+b\left[ab{x}^2+(ac+b^2)xy+bc{y}^2\right]$$

Expand each term:

1. $-c(ab{x}^2)=-abc{x}^2$
2. $-c(ac+b^2)xy=-a{c}^{2}xy-b^{2}cxy$
3. $-c(bc{y}^2)=-b{c}^2{y}^2$
4. $-a(b^2{x}^2)=-a{b}^2{x}^2$
5. $-a(2bcxy)=-2abcxy$
6. $-a(c^2{y}^2)=-a{c}^2{y}^2$
7. $b(ab{x}^2)=a{b}^2{x}^2$
8. $b(ac+b^2)xy=abcxy+b^{3}xy$
9. $b(bc{y}^2)=b^{2}c{y}^2$

Now combine all terms:

$$D=-abc{x}^2-a{c}^{2}xy-b^{2}cxy-b{c}^2{y}^2-a{b}^2{x}^2-2abcxy-a{c}^2{y}^2+a{b}^2{x}^2+abcxy+b^{3}xy+b^{2}c{y}^2$$

Simplify by canceling and combining like terms:

• $x^2$: $-abc{x}^2-a{b}^2{x}^2+a{b}^2{x}^2=-abc{x}^2$
• $xy$: $-a{c}^{2}xy-b^{2}cxy-2abcxy+abcxy+b^{3}xy=(-a{c}^2-b^{2}c-abc+b^3)xy$
• $y^2$: $-b{c}^2{y}^2-a{c}^2{y}^2+b^{2}c{y}^2=(-b{c}^2-a{c}^2+b^{2}c)y^2$

This still seems inconsistent. Instead, let’s use a smarter approach:

Notice that the determinant can be written as:

$$D=\left|\begin{array}{ccc}a& b& u\\ b& c& v\\ u& v& 0\end{array}\right|$$

where $u=ax+by$ and $v=bx+cy$.

The determinant of such a matrix is:

$$D=(b^2-ac)(u^2+v^2)-(bu-av{)}^2$$

But this seems incorrect. Instead, let’s use the formula for a 3×3 determinant:

$$D=a(c\cdot 0-v\cdot v)-b(b\cdot 0-v\cdot u)+(ax+by)(b\cdot v-c\cdot u)$$

$$=a(-v^2)-b(-uv)+(ax+by)(bv-cu)$$

$$=-a{v}^2+bvu+(ax+by)(bv-cu)$$

Substitute $u$ and $v$:

$$=-a(bx+cy{)}^2+b(ax+by)(bx+cy)+(ax+by)(b(bx+cy)-c(ax+by))$$

This is getting too complex. Instead, let’s verify the given identity directly:

We need to show:

$$D=(b^2-ac)(a{x}^2+2bxy+c{y}^2)$$

Let’s expand the right-hand side (RHS):

$$(b^2-ac)(a{x}^2+2bxy+c{y}^2)=b^2(a{x}^2+2bxy+c{y}^2)-ac(a{x}^2+2bxy+c{y}^2)$$

$$=a{b}^2{x}^2+2b^{3}xy+b^{2}c{y}^2-a^{2}c{x}^2-2abcxy-a{c}^2{y}^2$$

Now, let’s expand $D$ directly using the determinant formula:

$$D=a(c\cdot 0-(bx+cy{)}^2)-b(b\cdot 0-(bx+cy)(ax+by))+(ax+by)(b(bx+cy)-c(ax+by))$$

$$=-a(bx+cy{)}^2+b(ax+by)(bx+cy)+(ax+by)(b^{2}x+bcy-acx-bcy)$$

$$=-a(b^2{x}^2+2bcxy+c^2{y}^2)+b(ab{x}^2+(b^2+ac)xy+bc{y}^2)+(ax+by)(b^{2}x-acx)$$

$$=-a{b}^2{x}^2-2abcxy-a{c}^2{y}^2+a{b}^2{x}^2+(b^3+abc)xy+b^{2}c{y}^2+a{b}^2{x}^2-a^{2}c{x}^2$$

Combine like terms:

• $x^2$: $-a{b}^2{x}^2+a{b}^2{x}^2+a{b}^2{x}^2-a^{2}c{x}^2=(a{b}^2-a^{2}c)x^2$
• $xy$: $-2abcxy+(b^3+abc)xy=(-2abc+b^3+abc)xy=(b^3-abc)xy$
• $y^2$: $-a{c}^2{y}^2+b^{2}c{y}^2=(b^{2}c-a{c}^2)y^2$

This still does not match the RHS. Instead, let’s use a substitution approach:

Let $u=ax+by$ and $v=bx+cy$. Then:

$$D=\left|\begin{array}{ccc}a& b& u\\ b& c& v\\ u& v& 0\end{array}\right|$$

The determinant of such a matrix is:

$$D=-u^{2}c-v^{2}a+uvb+u^{2}b-u{v}^2+v^{2}b$$

This seems incorrect. Instead, let’s use the property of determinants for matrices of the form:

$$\left|\begin{array}{ccc}a& b& u\\ b& c& v\\ u& v& 0\end{array}\right|=(b^2-ac)(u^2+v^2)-(bu-av{)}^2$$

But this does not simplify to the given expression. Instead, let’s **re