Question

If $A$ is a square matrix of order $n$ and $A = kB$, where $k$ is a scalar, then $\left| A \right| = $
A. $\left| B \right|$
B. $k\left| B \right|$
C. ${k^n}\left| B \right|$
D. $n\left| B \right|$

Answer 1

Hint: In this question, we have to find that determinant of matrix $A$ is equal to which of the conditions given in the options. To find the answer to the given question, we must know the basic operations on matrices, such as addition, subtraction, multiplication, transpose, and determinant. In this question, we will assume a matrix $A$ and then from that, we will find $k$ and matrix $B$ and check which of the given options matches with the determinant of matrix $A$.

In order to find the determinant of \[2 \times 2\] matrix, the formula used is given below:
\[M = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right]\]
\[\left| M \right| = (({a_{11}}{a_{22}}) - ({a_{12}}{a_{21}}))\]

Complete step by step Solution:
Let us assume a matrix $\;A = \;\left[ {\begin{array}{*{20}{c}}
  2&4 \\
  6&8
\end{array}} \right]$ of order $(2 \times 2)$
On taking 2 commons from matrix $A$, it will be
$A = 2\;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right]$
It is given that $A = kB$, so from above, $k = 2$ and $B = \;\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right]$
Now, in order to find the determinant of matrix $A$, that is $\left| A \right|$, use the formula mentioned above:
For matrix$A$, $\;{a_{11}} = 2,{a_{12}} = 4,{a_{21}} = 6,{a_{22}} = 8$
So, $\;\left| A \right| = \;((2 \times 8) - (4 \times 6))$
     $\; = (16 - 24)$
$\;\left| A \right| = \; - 8 \ldots \ldots \ldots \ldots \ldots eq(1)$
Checking the condition of option A:
$\left| A \right| = \left| B \right|$ or not $?$
In order to find the transpose of $B$, that is $\left| B \right|$, use the formula mentioned above:
For matrix $B$, $\;{a_{11}} = 1,{a_{12}} = 2,{a_{21}} = 3,{a_{22}} = 4$
So, \[\;\left| B \right| = \;((1 \times 4) - (2 \times 3))\]
     $\; = (4 - 6)$
$\;\left| B \right| = \; - 2 \ldots \ldots \ldots \ldots \ldots eq(2)$
Here, from $eq(1)$ and $eq(2)$, it is clear that $\;\left| A \right| \ne \left| B \right|$
So, option A is not correct.

Checking the condition of option B:
$\left| A \right| = k\left| B \right|$ or not $?$
$k = 2$ and from $eq(2)$,$\;\left| B \right| = \; - 2$
So, $k\left| B \right| \Rightarrow 2( - 2) = - 4$
Here, from $eq(1)$ , it is clear that $\left| A \right| \ne k\left| B \right|$
So, option B is not correct.

Checking the condition of option C:
$\left| A \right| = {k^n}\left| B \right|$ or not $?$
$k = 2$ and from $eq(2)$,$\;\left| B \right| = \; - 2$
Now, ${k^n} \Rightarrow {2^2} = 4$ $(\because $matrix$A$ is of order 2 $)$
${k^n}\left| B \right| \Rightarrow (4 \times ( - 2)) = - 8$
Here, from $eq(1)$ , it is clear that $\left| A \right| = {k^n}\left| B \right|$
So, option C is correct.

Checking the condition of option D:
$\left| A \right| = n\left| B \right|$ or not $?$
$n = 2$ $(\because $matrix$A$is of order 2$)$
Now, $n\left| B \right| \Rightarrow (2 \times ( - 2)) = - 4$
Here, from $eq(1)$ , it is clear that $\left| A \right| \ne n\left| B \right|$
So, option D is not correct.

Therefore, the correct option is (C).

Additional Information: Since this question is based on the properties of matrices and mainly focuses on determinants of the matrix if we know the basic properties, then we can directly choose the correct option without assuming the matrices.

Note: To solve this type of question, where we are dealing with scalar multiplication of the matrix, always keep in mind that, scalar quantity is multiplied with every element of the matrix.