# Let $A$ be a square matrix of order $3 \times 3$, then $|kA|$ is equal to:

$(A)\;k|\;A\;|\qquad(B)\;k^2|\;A\;|\qquad(C)\;k^3|\;A\;|\qquad(D)\; 3k|\;A\;|$

Toolbox:
• If each element of a row (or column) of a determinant is multiplied by a constant k ,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column of the given determinant.
Let $A=\begin{vmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{vmatrix}$

We know $kA=\begin{vmatrix}ka_1 & kb_1 & kc_1\\ka_2 &k b_2 & kc_2\\ka_3 &k b_3 &k c_3\end{vmatrix}$

Then the value of the determinant is

$|kA|=\begin{vmatrix}ka_1 &k b_1 & kc_1\\ka_2 & kb_2 &k c_2\\ka_3 &k b_3 & kc_3\end{vmatrix}$

Now taking k as the common factor from $R_1,R_2,R_3$

$|kA|=k^3\begin{vmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{vmatrix}$

Therefore $|kA|=k^3|A|$

Hence C is the correct answer.

edited Feb 28, 2013