Number of real values λ , such that system of linear equations 2 x-3 y+5 z=9 ; x+3 y-z

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3 and 4 .Determinants and Matrices

hard

The number of real values $\lambda$, such that the system of linear equations $2 x-3 y+5 z=9$ ; $x+3 y-z=-18$ ; $3 x-y+\left(\lambda^{2}-1 \lambda \mid\right) z=16$ has no solution, is :-

$0$

$1$

$2$

$4$

(JEE MAIN-2022)

Solution

$\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda^{2}-|\lambda|\end{array}\right|=2\left(3 \lambda^{2}-3|\lambda|-1\right)$

$+3\left(\lambda^{2}-|\lambda|+3\right)$

$+5(-1-9)$

$=9 \lambda^{2}-9|\lambda|-43$

$=9|\lambda|^{2}-9|\lambda|-43$

$\Delta=0$ for 2 values of $|\lambda|$ out of which one is $-ve$ and other is $+ve$

So,$2$ values of $\lambda$ satisfy the system of equations to obtain no solution

Standard 12

Mathematics

For positive numbers $x,y$ and $z$ the numerical value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{{{\log }_x}y}&{{{\log }_x}z}\\{{{\log }_y}x}&1&{{{\log }_y}z}\\{{{\log }_z}x}&{{{\log }_z}y}&1\end{array}\,} \right|$is

easy

(IIT-1993)

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medium

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A&{{A^2}}&{ – B}\\
{{e^{A + B}}}&{{B^2}}&{ – 1}\\
1&{{A^2} + {B^2}}&{ – 1}
\end{array}} \right|$ is

normal

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medium

(IIT-2000)

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