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

$\left| {\,\begin{array}{*{20}{c}}{1/a}&1&{bc}\\{1/b}&1&{ca}\\{1/c}&1&{ab}\end{array}\,} \right| = $

easy

The value of a for which the system of equations ${a^3}x + {(a + 1)^3}y + {(a + 2)^3}z = 0,$ $ax + (a + 1)y + (a + 2)z = 0,$ $x + y + z = 0,$ has a non zero solution is

hard

If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions, then the value of $a$ is

easy

(IIT-2003)

If $1,\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\\{{\omega ^{2n}}}&1&{{\omega ^n}}\end{array}\,} \right|$ is equal to

medium

(AIEEE-2003)

Let $A =$ $\left[ {\begin{array}{*{20}{c}}{1 + {x^2} – {y^2} – {z^2}}&{2(xy + z)}&{2(zx – y)}\\{2(xy – z)}&{1 + {y^2} – {z^2} – {x^2}}&{2(yz + x)}\\{2(zx + y)}&{2(yz – x)}&{1 + {z^2} – {x^2} – {y^2}}\end{array}} \right]$ then det. $A$ is equal to