If | \beginmatrix a & a^2 & 1 + a^3 b & b^2 & 1 + b^3 c & c^2 & 1 + c^3 \endmatrix | = 0 and vectors ( 1,a,a^2 ),\,\,( 1,b,b^2 ) and ( 1,c,c^2 )\, are non-coplanar, then the product abc equals :

Vector Algebra PYQ — JEE Main 2003 · Shift 0 · Q96 with Solution

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Correct answer: C

a & a^2 & 1+a^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3 \end{vmatrix}=0.$$ We are also given that the vectors $$ (1,a,a^2),\ (1,b,b^2),\ (1,c,c^2) $$ are non-coplanar. For three vectors in $\mathbb R^3$, non-coplanar means they are linearly independent, so the determinant of the matrix formed by them is nonzero: $$\begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix} \neq 0.$$ 2. Evaluate the given determinant. Notice that $$1+a^3=1+a\cdot a^2,$$ and similarly for $b,c$. So the third column can be written as $$C_3=\begin{pmatrix}1+a^3\\1+b^3\\1+c^3\end{pmatrix} =\begin{pmatrix}1\\1\\1\end{pmatrix}+\begin{pmatrix}a^3\\b^3\\c^3\end{pmatrix}.

But

=\text{(entrywise product of first and second columns)}.$$ A better way is to relate the determinant directly to the Vandermonde determinant. Let us perform the column operation: $$C_3 \to C_3 - C_1C_2$$ which is not a valid linear column operation because $C_1C_2$ is not a column combination in the usual determinant sense. So instead we use row-wise polynomial reasoning. For each row corresponding to $x=a,b,c$, the row is $$(x,x^2,1+x^3).$$ Now observe that if we cyclically reorder columns as $$\begin{vmatrix} 1+a^3 & a & a^2\\ 1+b^3 & b & b^2\\ 1+c^3 & c & c^2 \end{vmatrix},$$ this only changes determinant by sign, which is not important for the zero condition. Now note that $$1+x^3=(x+1)(x^2-x+1).$$ But the cleanest route is to express the determinant as a linear combination of Vandermonde-type columns. Write $$1+x^3 = 1 + x^3.$$ So consider rows as values of functions $x, x^2, 1+x^3$ at $x=a,b,c$. The determinant vanishes iff these three functions are linearly dependent on the three distinct points $a,b,c$. Because the vectors $(1,a,a^2),(1,b,b^2),(1,c,c^2)$ are non-coplanar, $$\begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix}\neq 0.$$ This is the Vandermonde determinant: $$ (b-a)(c-a)(c-b)\neq 0,$$ so $a,b,c$ are distinct. 3. Since $a,b,c$ are distinct, there exists a unique quadratic polynomial $$p(x)=\alpha+\beta x+\gamma x^2$$ that matches the values $$p(a)=1+a^3,\quad p(b)=1+b^3,\quad p(c)=1+c^3.$$ Then the given determinant $$\begin{vmatrix} a & a^2 & 1+a^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3 \end{vmatrix}=0$$ means the third column is a linear combination of the first two columns and the vector $(1,1,1)^T$. Equivalently, there exist constants $\alpha,\beta,\gamma$, not all zero, such that $$1+x^3=\alpha x+\beta x^2+\gamma$$ for $x=a,b,c$. Rearranging, $$x^3-\beta x^2-\alpha x+(1-\gamma)=0$$ for $x=a,b,c$. Thus $a,b,c$ are the three roots of a cubic polynomial of the form $$x^3-\beta x^2-\alpha x+(1-\gamma).$$ But from the determinant form, the columns are $x, x^2, 1+x^3$, so linear dependence of these three columns means there exist constants $\lambda,\mu,\nu$, not all zero, such that for $x=a,b,c$, $$\lambda x+\mu x^2+\nu(1+x^3)=0.$$ If $\nu=0$, then $\lambda x+\mu x^2=0$ at three distinct values, impossible unless $\lambda=\mu=0$. Hence $\nu\neq 0$. Divide by $\nu$: $$x^3+\frac{\mu}{\nu}x^2+\frac{\lambda}{\nu}x+1=0$$ for $x=a,b,c$. So $a,b,c$ are roots of a monic cubic $$x^3+px^2+qx+1=0.$$ 4. By Vieta's formulas, for a monic cubic $$x^3+px^2+qx+1=0,$$ the product of roots is $$abc=-1.$$ Therefore, $$\boxed{abc=-1}.$$ 5. Checking options: - A: $0$ - B: $2$ - C: $-1$ ✅ - D: $1$ So the correct option is $$\boxed{\text{C}}.$$

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View written solutionFree

Correct answer: C

a & a^2 & 1+a^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3 \end{vmatrix}=0.$$ We are also given that the vectors $$ (1,a,a^2),\ (1,b,b^2),\ (1,c,c^2) $$ are non-coplanar. For three vectors in $\mathbb R^3$, non-coplanar means they are linearly independent, so the determinant of the matrix formed by them is nonzero: $$\begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix} \neq 0.$$ 2. Evaluate the given determinant. Notice that $$1+a^3=1+a\cdot a^2,$$ and similarly for $b,c$. So the third column can be written as $$C_3=\begin{pmatrix}1+a^3\\1+b^3\\1+c^3\end{pmatrix} =\begin{pmatrix}1\\1\\1\end{pmatrix}+\begin{pmatrix}a^3\\b^3\\c^3\end{pmatrix}.

But

=\text{(entrywise product of first and second columns)}.$$ A better way is to relate the determinant directly to the Vandermonde determinant. Let us perform the column operation: $$C_3 \to C_3 - C_1C_2$$ which is not a valid linear column operation because $C_1C_2$ is not a column combination in the usual determinant sense. So instead we use row-wise polynomial reasoning. For each row corresponding to $x=a,b,c$, the row is $$(x,x^2,1+x^3).$$ Now observe that if we cyclically reorder columns as $$\begin{vmatrix} 1+a^3 & a & a^2\\ 1+b^3 & b & b^2\\ 1+c^3 & c & c^2 \end{vmatrix},$$ this only changes determinant by sign, which is not important for the zero condition. Now note that $$1+x^3=(x+1)(x^2-x+1).$$ But the cleanest route is to express the determinant as a linear combination of Vandermonde-type columns. Write $$1+x^3 = 1 + x^3.$$ So consider rows as values of functions $x, x^2, 1+x^3$ at $x=a,b,c$. The determinant vanishes iff these three functions are linearly dependent on the three distinct points $a,b,c$. Because the vectors $(1,a,a^2),(1,b,b^2),(1,c,c^2)$ are non-coplanar, $$\begin{vmatrix} 1 & a & a^2\\ 1 & b & b^2\\ 1 & c & c^2 \end{vmatrix}\neq 0.$$ This is the Vandermonde determinant: $$ (b-a)(c-a)(c-b)\neq 0,$$ so $a,b,c$ are distinct. 3. Since $a,b,c$ are distinct, there exists a unique quadratic polynomial $$p(x)=\alpha+\beta x+\gamma x^2$$ that matches the values $$p(a)=1+a^3,\quad p(b)=1+b^3,\quad p(c)=1+c^3.$$ Then the given determinant $$\begin{vmatrix} a & a^2 & 1+a^3\\ b & b^2 & 1+b^3\\ c & c^2 & 1+c^3 \end{vmatrix}=0$$ means the third column is a linear combination of the first two columns and the vector $(1,1,1)^T$. Equivalently, there exist constants $\alpha,\beta,\gamma$, not all zero, such that $$1+x^3=\alpha x+\beta x^2+\gamma$$ for $x=a,b,c$. Rearranging, $$x^3-\beta x^2-\alpha x+(1-\gamma)=0$$ for $x=a,b,c$. Thus $a,b,c$ are the three roots of a cubic polynomial of the form $$x^3-\beta x^2-\alpha x+(1-\gamma).$$ But from the determinant form, the columns are $x, x^2, 1+x^3$, so linear dependence of these three columns means there exist constants $\lambda,\mu,\nu$, not all zero, such that for $x=a,b,c$, $$\lambda x+\mu x^2+\nu(1+x^3)=0.$$ If $\nu=0$, then $\lambda x+\mu x^2=0$ at three distinct values, impossible unless $\lambda=\mu=0$. Hence $\nu\neq 0$. Divide by $\nu$: $$x^3+\frac{\mu}{\nu}x^2+\frac{\lambda}{\nu}x+1=0$$ for $x=a,b,c$. So $a,b,c$ are roots of a monic cubic $$x^3+px^2+qx+1=0.$$ 4. By Vieta's formulas, for a monic cubic $$x^3+px^2+qx+1=0,$$ the product of roots is $$abc=-1.$$ Therefore, $$\boxed{abc=-1}.$$ 5. Checking options: - A: $0$ - B: $2$ - C: $-1$ ✅ - D: $1$ So the correct option is $$\boxed{\text{C}}.$$

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