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

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

hard

(IIT-2012)

If $a, b, c > 0 \, and \, x, y, z \in R$ , then the determinant $\left|{\begin{array}{*{20}{c}}{{{\left( {{a^x}\, + \,\,{a^{ – x}}} \right)}^2}}&{{{\left( {{a^x}\, – \,\,{a^{ – x}}} \right)}^2}}&1\\{{{\left( {{b^y}\, + \,\,{b^{ – y}}} \right)}^2}}&{{{\left( {{b^y}\, – \,\,{b^{ – y}}} \right)}^2}}&1\\{{{\left( {{c^z}\, + \,\,{c^{ – z}}} \right)}^2}}&{{{\left( {{c^z}\, – \,\,{c^{ – z}}} \right)}^2}}&1\end{array}} \right|$ $=$

normal

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to …… .

medium

(JEE MAIN-2021)

Solve the equations $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0, a \neq 0$

medium

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

medium

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

Let $P=\left[a_{\|}\right]$be $a \times 3$ matrix and let $Q=\left[b_1\right]$, where $b_1=2^{1+j} a_{\|}$for $1 \leq i, j \leq 3$. If the determinant of $P$ is $2$ , then the determinant of the matrix $Q$ is

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}( A )=4$. Let $R _{ i }$ denote the $i ^{\text {th }}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R _{2} \rightarrow 2 R _{2}+5 R _{3}$ on $2 A ,$ then $\operatorname{det}( B )$ is equal to …… .

Solve the equations $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0, a \neq 0$

Using the property of determinants and without expanding, prove that: $\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$

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Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$