If x, y, z are different and Delta=left|begin{array}{lll}x & x^{2} & 1+x^{2} y & y^{2} &

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3 and 4 .Determinants and Matrices

hard

If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.

Option A

Option B

Option C

Option D

Solution

We have $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|$

$=\left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right|+\left|\begin{array}{lll}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|$

$=(-1)^{2}\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+x y z\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \quad\left(\text { Using } \mathrm{C}_{3} \leftrightarrow \mathrm{C}_{2} \text { and then } \mathrm{C}_{1} \leftrightarrow \mathrm{C}_{2}\right)$

$=\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|(1+x y z)$

$=(1+x y z)\left|\begin{array}{ccc}
1 & x & x^{2} \\
0 & y-x & y^{2}-x^{2} \\
0 & z-x & z^{2}-x^{2}
\end{array}\right| $ $( { Using } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1})$

Taking out common factor $(y-x)$ from $\mathrm{R}_{2}$ and $(z-x)$ from $\mathrm{R}_{3},$ we get

$\Delta=(1+x y z)(y-x)(z-x)\left|\begin{array}{ccc}
1 & x & x^{2} \\
0 & 1 & y+x \\
0 & 1 & z+x
\end{array}\right|$

$=(1+x y z)(y-x)(z-x)(z-y)$ (on expanding along $\mathrm{C}_{1}$ )

since $\Delta=0$ and $x, y, z$ are all different, i.e., $x-y \neq 0, y-z \neq 0, z-x \neq 0,$ we get $1+x y z=0$

Standard 12

Mathematics

If ${a_1},{a_2},{a_3},……..,{a_n},……$ are in G.P. and ${a_i} > 0$ for each $i$, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 2}}}&{\log {a_{n + 4}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 8}}}&{\log {a_{n + 10}}}\\{\log {a_{n + 12}}}&{\log {a_{n + 14}}}&{\log {a_{n + 16}}}\end{array}} \right|$ is equal to

medium

Value of $\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ is equal to

normal

The value of $\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ is equal to

medium

If $\omega $ is a complex cube root of unity, then the determinant $\left| {\,\begin{array}{*{20}{c}}2&{2\omega }&{ – {\omega ^2}}\\1&1&1\\1&{ – 1}&0\end{array}\,} \right| = $

easy

If $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x – 1)}&{(x + 1)x}\\{3x(x – 1)}&{x(x – 1)(x – 2)}&{(x + 1)x(x – 1)}\end{array}} \right|$ then $f(100)$ is equal to

hard

(IIT-1999)

If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.

Solution

Similar Questions

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The value of $\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ is equal to

If $\omega $ is a complex cube root of unity, then the determinant $\left| {\,\begin{array}{*{20}{c}}2&{2\omega }&{ – {\omega ^2}}\\1&1&1\\1&{ – 1}&0\end{array}\,} \right| = $

If $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x – 1)}&{(x + 1)x}\\{3x(x – 1)}&{x(x – 1)(x – 2)}&{(x + 1)x(x – 1)}\end{array}} \right|$ then $f(100)$ is equal to

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Value of $\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ is equal to