Three digit numbers x17, 3y6 and 12z where x, y, z are integers from 0 to 9 , are divisible by a fix

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3 and 4 .Determinants and Matrices

normal

Three digit numbers $x17, 3y6$ and $12z$ where $x, y, z$ are integers from $0$ to $9$, are divisible by a fixed constant $k$. Then the determinant $\left| {\,\begin{array}{*{20}{c}}x&3&1\\7&6&z\\1&y&2\end{array}\,} \right|$ + $48$ must be divisible by

$k$

$k^2$

$k^3$

None

Solution

since $7 x, 36 y, 12 z$ are divisible by $k$ Let us Assume $x=k a, y=k b, z=k c$ So the det $\Longrightarrow\left|\begin{array}{ccc}k a & 3 & 1 \\ 7 & 6 & k c \\ 1 & k b & 2\end{array}\right|$

$\Longrightarrow 12 k a-k^{3} a b c+3 k c+7 k b-48$

$\Longrightarrow$ From Question, $12 \mathrm{ka}-\mathrm{k}^{3} \mathrm{abc}+3 \mathrm{kc}+7 \mathrm{kb}-48+48$

Hence A

Standard 12

Mathematics

If the system of linear equations $2 \mathrm{x}+2 \mathrm{ay}+\mathrm{az}=0$ ; $2 x+3 b y+b z=0$ ; $2 \mathrm{x}+4 \mathrm{cy}+\mathrm{cz}=0$ ; where $a, b, c \in R$ are non-zero and distinct; has a non-zero solution, then

hard

(JEE MAIN-2020)

For what value of $\lambda $, the system of equations $x + y + z = 6,x + 2y + 3z = 10,$ $x + 2y + \lambda z = 12$ is inconsistent $\lambda =$ ……..

medium

(AIEEE-2002)

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easy

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$x+y+z=2$

$2 x+4 y-z=6$

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