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Scalar Matrix

If x i = a ...

Question

If $x_{i} = a_{i} b_{i} c_{i}, i = 1, 2, 3,$ are theree-digit positive integers such that each $x_{i}$ is a multiple of $19,$ then for some integer n,
$△ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{2} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ is given by

19 n + 1

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19 n + 2

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19 n

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19 n + 3

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Solution

The correct option is C $19 n$
$△ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} (100 a_{1} + 10 b_{1} + C_{1}) & (100 a_{2} + 10 b_{2} + C_{2}) & (100 a_{3} + 10 b_{3} + C_{3}) \end{matrix} ∣ ∣ ∣ ∣$
$($ using $R_{3} \to R_{3} + 100 R_{1} + 10 R_{2})$
$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} x_{1} & x_{2} & x_{3} \end{matrix} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} {19 m}_{1} & {19 m}_{2} & {19 m}_{3} \end{matrix} ∣ ∣ ∣ ∣$
$($ where each $m_{i} \in N)$
$= 19 ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} m_{1} & m_{2} & m_{3} \end{matrix} ∣ ∣ ∣ ∣ = 19 n$
where $n = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} m_{1} & m_{2} & m_{3} \end{matrix} ∣ ∣ ∣ ∣$ is certainly an integer.

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Similar questions

x_{i} = a_{i} b_{i} c_{i}

i = 1, 2, 3

x_{i}

19

n

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

19

x_{1} = a_{i} b_{i} c_{i}, i = 1, 2, 3

x_{1}

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∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

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Check whether the value given in the brackets is a solution to the given equation or not:

(a) n + 5 = 19 (n = 1) (b) 7n + 5 = 19 (n = − 2)

(e) 4p − 3 = 13 (p = − 4) (f) 4p − 3 = 13 (p = 0)

n

n^{2} + 19 n + 92

Solve: $n + 5 = 19 (n - 1)$

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