Question

Assertion: The determinant of a skew symmetric matrix of even order is perfect square.
Reason: The determinant of skew symmetric matrix of odd order is equal to zero
(a) Assertion and reason both are correct and reason is correct explanation of assertion
(b) Assertion and reason both are correct but reason is not correct explanation of assertion
(c) Assertion is wrong
(d) Reason is wrong

Answer 1

Hint: We solve this problem by checking for assertion and reason one by one.
For assertion, we take an example of a skew-symmetric matrix of even order and check the determinant.
For reason, we use some conditions
(1) Any matrix is said to be skew-symmetric if and only if
\[{{A}^{T}}=-A\]
Where \[{{A}^{T}}\] is called transpose of \[A\]
(2) The determinant of transpose
\[\left| {{A}^{T}} \right|=\left| A \right|\]
(3) Determinant of scalar multiple of matrix
\[\left| kA \right|={{k}^{n}}\left| A \right|\]
Where \[k\] is real number and \[n\] is order of matrix

Complete step by step answer:
We are given two statements of assertion and reason.
Let us check the assertion first which is given as the determinant of a skew-symmetric matrix of even order is a perfect square.
Let us assume a skew symmetric matrix of order 2 as follows
\[\Rightarrow A=\left[ \begin{matrix}
   0 & -a \\
   a & 0 \\
\end{matrix} \right]\]
Where \[a\in \mathbb{R}\]
Now, let us find the determinant of the above matrix then we get
\[\begin{align}
  & \Rightarrow \left| A \right|=\left| \begin{matrix}
   0 & -a \\
   a & 0 \\
\end{matrix} \right| \\
 & \Rightarrow \left| A \right|=0-\left( -{{a}^{2}} \right) \\
 & \Rightarrow \left| A \right|={{a}^{2}} \\
\end{align}\]
Here, we can see that the determinant of matrix is \[{{a}^{2}}\]
Here, we can say that determinant of matrix is perfect square if and only if \[a\in \mathbb{N}\cup \left\{ 0 \right\}\]
We know that the value of \[a\] is a real number that is \[a\in \mathbb{R}\]
Here, we can see that there are some values of \[a\] for which the value \[{{a}^{2}}\] is not perfect square for example \[a=\sqrt{3}\]
Therefore we can conclude that the assertion is wrong.
Let us check the reason that is the determinant of skew symmetric matrix of odd order is equal to zero
Let us assume that the matrix \[A\] of odd order as \[n\]
We know that the condition for skew symmetric matrix given as
\[{{A}^{T}}=-A\]
Where \[{{A}^{T}}\] is called transpose of \[A\]
Now, let us apply the determinant for the above condition then we get
\[\Rightarrow \left| {{A}^{T}} \right|=\left| -A \right|\]
We know that the formulas of determinant of transpose and determinant of scalar multiple of a matrix given as
\[\left| {{A}^{T}} \right|=\left| A \right|\] and \[\left| kA \right|={{k}^{n}}\left| A \right|\]
By using the these formula to above equation we get
\[\Rightarrow \left| A \right|={{\left( -1 \right)}^{n}}\left| A \right|\]
We know that \[n\] is odd such that \[{{\left( -1 \right)}^{n}}=-1\]
By using this condition to above equation we get
\[\begin{align}
  & \Rightarrow \left| A \right|=-\left| A \right| \\
 & \Rightarrow 2\left| A \right|=0 \\
 & \Rightarrow \left| A \right|=0 \\
\end{align}\]
Here, we can see that the determinant of skew symmetric matrix of odd order is 0
Therefore we can conclude that the reason is correct.
Hence, the assertion is wrong and the reason is correct
So, option (c) is the correct answer.

Note:
We need to note that in the determinant of skew-symmetric matrix of even order we have the determinant as
\[\Rightarrow \left| A \right|={{a}^{2}}\]
Here, some students may directly give the answer that \[{{a}^{2}}\] is a perfect square.
But we need to confirm with the domain of \[a\] that is \[a\in \mathbb{R}\]
Here, for all \[a\in \mathbb{N}\cup \left\{ 0 \right\}\]the determinant will be a perfect square but there are other numbers in the real numbers where the determinant is not a perfect square.
We need to check for all possibilities.