Question

If \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  m&n&p \\
  x&y&z
\end{array}} \right| = k\], then \[\left| {\begin{array}{*{20}{c}}
  {6a}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right| = \]
A. \[\dfrac{k}{6}\]
B. \[2k\]
C. \[3k\]
D. \[6k\]

Answer 1

Hint: By using the properties of determinants, first of all take the common term in the first column and then take the common term in the first row of the determinant. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  m&n&p \\
  x&y&z
\end{array}} \right| = k............................................\left( 1 \right)\]
Now consider \[\left| {\begin{array}{*{20}{c}}
  {6a}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right|\]
We know if there is a common term in any row or column of a determinant, then we can take it common by multiplying the common terms with the determinant.
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {6a}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {3\left( {2a} \right)}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right|\]
In the first column i.e., \[{C_1}\] we can take 3 as a common term. So, we have
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {6a}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right| = 3\left| {\begin{array}{*{20}{c}}
  {2a}&{2b}&{2c} \\
  m&n&p \\
  x&y&z
\end{array}} \right|\]
In the first row i.e., \[{R_1}\] we can take 2 as a common term. So, we have
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {6a}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right| = 3 \times 2\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  m&n&p \\
  x&y&z
\end{array}} \right| = 6\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  m&n&p \\
  x&y&z
\end{array}} \right|..................................................\left( 2 \right)\]
From equation (1) and (2), we have
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {6a}&{2b}&{2c} \\
  {3m}&n&p \\
  {3x}&y&z
\end{array}} \right| = 6k\]
Thus, the correct option is D. \[6k\]

Note: If there is a common term in any row or column of a determinant, then we can take it common by multiplying the common terms with the determinant. This property is also known as scalar multiple property of the determinant.