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3 and 4 .Determinants and Matrices
medium
समीकरण निकाय $kx + y + z =1, x + ky + z = k$ तथा $x + y + zk = k ^{2}$ का कोई हल नहीं है, यदि $k$ बराबर है
A
$0$
B
$1$
C
$-1$
D
$-2$
(JEE MAIN-2021)
Solution
$k x+y+z=1$
$x+k y+z=k$
$x+y+z k=k^{2}$
$\Delta=\left|\begin{array}{ccc} K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K \end{array}\right|= K \left( K ^{2}-1\right)-1( K -1)+1(1- K )$
$= K ^{3}- K – K +1+1- K$
$= K ^{3}-3 K +2$
$=( K -1)^{2}( K +2)$
For $K =1$
$\Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$
But for $K =-2,$ at least one out of $\Delta_{1}, \Delta_{2}, \Delta_{3}$ are not zero
Hence for no sol$^{n}$, $K =-2$
Standard 12
Mathematics
