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Evaluation of a Determinant

Let |A|=|aij|...

Question

Let $| A | = | a_{i j} |_{3 \times 3} \neq 0$ . Each element $a_{i j}$ multiplied by $k^{i - j}$ . Let |B| be the resulting determinant, where $k_{1} | A | + k_{2} | B | = 0. T h e n k_{1} + k_{2} =$

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Solution

The correct option is C
0

$| A | = ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣ | B | = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & k^{- 1} a_{12} & k^{- 2} a_{13} k a_{21} & a_{22} & k^{- 1} a_{23} k^{2} a_{31} & k a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{1}{k^{3}} ∣ ∣ ∣ ∣ ∣ \begin{matrix} k^{2} a_{11} & k a_{12} & a_{13} k^{2} a_{21} & k a_{22} & a_{23} k^{2} a_{31} & k a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{1}{k^{3}} \times k^{2} \times k ∣ ∣ ∣ ∣ \begin{matrix} a_{11} & a_{12} & a_{13} a_{21} & a_{22} & a_{23} a_{31} & a_{32} & a_{33} \end{matrix} ∣ ∣ ∣ ∣ = | A | k_{1} | A | + k_{2} | B | = 0 \Rightarrow | A | (k_{1} + k_{2}) = 0 \Rightarrow k_{1} + k_{2} = 0$

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Similar questions

Let $| A | = | a_{i j} |_{3 \times 3} \neq 0$ . Each element $a_{i j}$ multiplied by $k^{i - j}$ . Let |B| be the resulting determinant, where $k_{1} | A | + k_{2} | B | = 0. T h e n k_{1} + k_{2} =$

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Q. let

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Let |A|=|aij|3×3≠0. Each element aij multiplied by ki−j. Let |B| be the resulting determinant, where k1|A|+k2|B|=0. Then k1+k2=

Let $| A | = | a_{i j} |_{3 \times 3} \neq 0$ . Each element $a_{i j}$ multiplied by $k^{i - j}$ . Let |B| be the resulting determinant, where $k_{1} | A | + k_{2} | B | = 0. T h e n k_{1} + k_{2} =$