Equation left| {begin{array}{*{20}{c}}{{{(1 + x)}^2}}&{{{(1 - x)}^2}}&{

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3 and 4 .Determinants and Matrices

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The equation $\left| {\begin{array}{{20}{c}}{{{(1 + x)}^2}}&{{{(1 - x)}^2}}&{ - \,(2 + {x^2})}\\{2x + 1}&{3x}&{1 - 5x}\\{x + 1}&{2x}&{2 - 3x}\end{array}} \right|$ $+$ $\left| {\begin{array}{{20}{c}}{{{(1 + x)}^2}}&{2x + 1}&{x + 1}\\{{{(1 - x)}^2}}&{3x}&{2x}\\{1 - 2x}&{3x - 2}&{2x - 3}\end{array}} \right|$ $= 0$

Ahas no real solution

Bhas $4$ real solutions

Chas two real and two non-real solutions

Dhas infinite number of solutions , real or non-real

Solution

$1^{st}$ two columns of $1^{st}$ determinant are same as $1^{st}$ two rows of $2^{nd}$. Hence transpose the $2^{nd}$. Add the two determinants and use $C_1 \rightarrow C_1 + C_3$ ==> $D = 0$

Standard 12

Mathematics

If the system of equations

$ 2 x+7 y+\lambda z=3 $

$ 3 x+2 y+5 z=4 $

$ x+\mu y+32 z=-1$

has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

hard

(JEE MAIN-2024)

If $\left| {\,\begin{array}{*{20}{c}}a&b&0\\0&a&b\\b&0&a\end{array}\,} \right| = 0$, then

easy

If the system of equations $x +y + z = 6$ ; $x + 2y + 3z= 10$ ; $x + 2y + \lambda z = 0$ has a unique solution, then $\lambda $ is not equal to

hard

(AIEEE-2012)

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

hard

(JEE MAIN-2019)

If $a, b, c$ are sides of a scalene triangle, then the value of $\left| \begin{array}{*{20}{c}}
a&b&c\\
b&c&a\\
c&a&b
\end{array} \right|$ is

hard

(JEE MAIN-2013)

Solution

Similar Questions

If the system of equations $ 2 x+7 y+\lambda z=3 $ $ 3 x+2 y+5 z=4 $ $ x+\mu y+32 z=-1$ has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

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If the system of equations

$ 2 x+7 y+\lambda z=3 $

$ 3 x+2 y+5 z=4 $

$ x+\mu y+32 z=-1$

has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

If $a, b, c$ are sides of a scalene triangle, then the value of $\left| \begin{array}{*{20}{c}}
a&b&c\\
b&c&a\\
c&a&b
\end{array} \right|$ is