A - begin vmatrix k a - k b - k c l -kd - ke - kfllkg - kh - ki lend vmatrix B - begin vmatrix a - b - cd - e - flg - h - ilend vmatrix Whats the value of A - B - -A- k 3B- 1C- kD- 0

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Cofactor

A = begin vma...

Question

A = $∣ ∣ ∣ ∣ \begin{matrix} k a & k b & k c k d & k e & k f k g & k h & k i \end{matrix} ∣ ∣ ∣ ∣$
B = $∣ ∣ ∣ ∣ \begin{matrix} a & b & c d & e & f g & h & i \end{matrix} ∣ ∣ ∣ ∣$

Whats the value of A/B = ?

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k³

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Solution

The correct option is C
k³

t’s the property of the determinant that a scalar multiplied to a determinant will be same as the same scalar being multiplied to elements of one of the row or column of the same determinant. So if the scalar is multiplied to 3 rows or columns you can take out 3 times the scalar outside as below.

A = $∣ ∣ ∣ ∣ \begin{matrix} k a & k b & k c k d & k e & k f k g & k h & k i \end{matrix} ∣ ∣ ∣ ∣$

= $k^{3}$ $∣ ∣ ∣ ∣ \begin{matrix} a & b & c d & e & f g & h & i \end{matrix} ∣ ∣ ∣ ∣$

Therefore, A/B = $k^{3}$

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Similar questions

$If ∣ ∣ ∣ ∣ \begin{matrix} k a & k b & k c k d & k e & k f k g & k h & k i \end{matrix} ∣ ∣ ∣ ∣ = A and$
$∣ ∣ ∣ ∣ \begin{matrix} a & b & c d & e & f g & h & i \end{matrix} ∣ ∣ ∣ ∣ = B$
Then the value of $A / B$ is

If value of $∣ ∣ ∣ ∣ \begin{matrix} a & b & c d & e & f g & h & i \end{matrix} ∣ ∣ ∣ ∣$ is A then value of $∣ ∣ ∣ ∣ \begin{matrix} d & e & f a & b & c g & h & i \end{matrix} ∣ ∣ ∣ ∣$ is also equal to A.