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If $A$ is a matrix of order $3$, then det($kA$)

\[\begin{array}{1 1}(1)k^{3}det(A)&(2)k^{2}det(A)\\(3)k\;det (A)& (4)det(A)\end{array}\]

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Let $A= \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
$K A= \begin{bmatrix} Ka_{11} & Ka_{12} & Ka_{13} \\ Ka_{21} & Ka_{22} & Ka_{23} \\ Ka_{31} & Ka_{32} & Ka_{33} \end{bmatrix}$
$| K A |= \begin{vmatrix} Ka_{11} & Ka_{12} & Ka_{13} \\ Ka_{21} & Ka_{22} & Ka_{23} \\ Ka_{31} & Ka_{32} & Ka_{33} \end{vmatrix}$
$\qquad=K.K.K \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$
$\qquad= K^3 |A|$
Hence 1 is the correct answer.
answered May 5, 2014 by meena.p
 

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