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3 and 4 .Determinants and Matrices
hard
$\lambda$ के वास्तविक मानों, जिनके लिए रैखिक समीकरण निकाय
$2 x -3 y +5 z =9$
$x +3 y - z =-18$
$3 x - y +\left(\lambda^2-|\lambda|\right) z =16$
का कोई हल नहीं है, की संख्या है :-
A
$0$
B
$1$
C
$2$
D
$4$
(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
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