The total number of distinct x \in R for which | \beginmatrix x & x^2 & 1 + x^3 2x & 4x^2 & 1 + 8x^3 3x & 9x^2 & 1 + 27x^3 \endmatrix | = 10 is \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Advanced 2016 · Shift 1 · Q34 with Solution

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Correct answer: 2

x & x^2 & 1+x^3\\ 2x & 4x^2 & 1+8x^3\\ 3x & 9x^2 & 1+27x^3 \end{vmatrix}.$$ We need the number of real $x$ such that $$D=10.$$ 2. Use linearity of determinant in the third column: $$C_3=\begin{pmatrix}1+x^3\\1+8x^3\\1+27x^3\end{pmatrix} =\begin{pmatrix}1\\1\\1\end{pmatrix}+x^3\begin{pmatrix}1\\8\\27\end{pmatrix}.$$ So $$D= \begin{vmatrix} x & x^2 & 1\\ 2x & 4x^2 & 1\\ 3x & 9x^2 & 1 \end{vmatrix} +x^3 \begin{vmatrix} x & x^2 & 1\\ 2x & 4x^2 & 8\\ 3x & 9x^2 & 27 \end{vmatrix}.

Factor powers of $x$ from the first two columns. In the original determinant, column 1 has a factor $x$ and column 2 has a factor $x^2$ . Hence overall,

1 & 1 & 1+x^3\\ 2 & 4 & 1+8x^3\\ 3 & 9 & 1+27x^3 \end{vmatrix}.$$ Now split again in the third column: $$D=x^3\left[ \begin{vmatrix} 1 & 1 & 1\\ 2 & 4 & 1\\ 3 & 9 & 1 \end{vmatrix} +x^3 \begin{vmatrix} 1 & 1 & 1\\ 2 & 4 & 8\\ 3 & 9 & 27 \end{vmatrix} \right].$$ 4. Compute the two constant determinants. First, $$A=\begin{vmatrix} 1 & 1 & 1\\ 2 & 4 & 1\\ 3 & 9 & 1 \end{vmatrix}.$$ Apply row operations $R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$: $$A=\begin{vmatrix} 1 & 1 & 1\\ 0 & 2 & -1\\ 0 & 6 & -2 \end{vmatrix}.

Thus $A=1\cdot\begin{vmatrix}2 & -1\\6 & -2\end{vmatrix}=2(-2)-(-1)6=-4+6=2.$

Second,

1 & 1 & 1\\ 2 & 4 & 8\\ 3 & 9 & 27 \end{vmatrix}.$$ This is a Vandermonde-type determinant for $1,2,3$, so $$B=(2-1)(3-1)(3-2)=2.$$ (Direct expansion also gives $2$.) Hence $$D=x^3(A+x^3B)=x^3(2+2x^3)=2x^3(1+x^3).$$ 5. Now solve $$2x^3(1+x^3)=10.$$ So $$x^3(1+x^3)=5.$$ Let $$y=x^3.$$ Then $$y(y+1)=5,$$ which gives $$y^2+y-5=0.$$ So $$y=\frac{-1\pm\sqrt{21}}{2}.$$ 6. Each real value of $y$ gives exactly one real value of $x$, since $x=\sqrt[3]{y}$ is unique for every real $y$. Therefore, there are $$2$$ distinct real values of $x$. 7. Comparison with stored answer: Stored correct answer = $2$, which matches our result.

View written solutionFree

Correct answer: 2

x & x^2 & 1+x^3\\ 2x & 4x^2 & 1+8x^3\\ 3x & 9x^2 & 1+27x^3 \end{vmatrix}.$$ We need the number of real $x$ such that $$D=10.$$ 2. Use linearity of determinant in the third column: $$C_3=\begin{pmatrix}1+x^3\\1+8x^3\\1+27x^3\end{pmatrix} =\begin{pmatrix}1\\1\\1\end{pmatrix}+x^3\begin{pmatrix}1\\8\\27\end{pmatrix}.$$ So $$D= \begin{vmatrix} x & x^2 & 1\\ 2x & 4x^2 & 1\\ 3x & 9x^2 & 1 \end{vmatrix} +x^3 \begin{vmatrix} x & x^2 & 1\\ 2x & 4x^2 & 8\\ 3x & 9x^2 & 27 \end{vmatrix}.

Factor powers of $x$ from the first two columns. In the original determinant, column 1 has a factor $x$ and column 2 has a factor $x^2$ . Hence overall,

1 & 1 & 1+x^3\\ 2 & 4 & 1+8x^3\\ 3 & 9 & 1+27x^3 \end{vmatrix}.$$ Now split again in the third column: $$D=x^3\left[ \begin{vmatrix} 1 & 1 & 1\\ 2 & 4 & 1\\ 3 & 9 & 1 \end{vmatrix} +x^3 \begin{vmatrix} 1 & 1 & 1\\ 2 & 4 & 8\\ 3 & 9 & 27 \end{vmatrix} \right].$$ 4. Compute the two constant determinants. First, $$A=\begin{vmatrix} 1 & 1 & 1\\ 2 & 4 & 1\\ 3 & 9 & 1 \end{vmatrix}.$$ Apply row operations $R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$: $$A=\begin{vmatrix} 1 & 1 & 1\\ 0 & 2 & -1\\ 0 & 6 & -2 \end{vmatrix}.

Thus $A=1\cdot\begin{vmatrix}2 & -1\\6 & -2\end{vmatrix}=2(-2)-(-1)6=-4+6=2.$

Second,

1 & 1 & 1\\ 2 & 4 & 8\\ 3 & 9 & 27 \end{vmatrix}.$$ This is a Vandermonde-type determinant for $1,2,3$, so $$B=(2-1)(3-1)(3-2)=2.$$ (Direct expansion also gives $2$.) Hence $$D=x^3(A+x^3B)=x^3(2+2x^3)=2x^3(1+x^3).$$ 5. Now solve $$2x^3(1+x^3)=10.$$ So $$x^3(1+x^3)=5.$$ Let $$y=x^3.$$ Then $$y(y+1)=5,$$ which gives $$y^2+y-5=0.$$ So $$y=\frac{-1\pm\sqrt{21}}{2}.$$ 6. Each real value of $y$ gives exactly one real value of $x$, since $x=\sqrt[3]{y}$ is unique for every real $y$. Therefore, there are $$2$$ distinct real values of $x$. 7. Comparison with stored answer: Stored correct answer = $2$, which matches our result.