Question

A and B are square matrices of order $ 3\times 3 $ , A is an orthogonal matrix and B is a skew symmetric matrix. Which of the following statements is not true.
 $ \begin{align}
  & a.\left| A \right|=\pm 1 \\
 & b.\left| B \right|=0 \\
 & c.\left| AB \right|=1 \\
 & d.\left| AB \right|=0 \\
\end{align} $

Answer 1

Hint: We will refer to the basic definition of orthogonal matrix and skew symmetric matrix and check all the options one by one. We will also use the concept that for orthogonal matrix $ A{{A}^{T}}=I $ and for skew symmetric matrix B, we have $ {{B}^{T}}=-B $ .

Complete step-by-step answer:
It is given in the question that A and B are square matrices of order $ 3\times 3 $ , A is an orthogonal matrix and B is a skew symmetric matrix, then we have been asked to find which of the given statements is not true.
We know that the matrix A is said to be an orthogonal matrix if the product of the matrix and its transpose gives us the identity matrix, that is, $ A{{A}^{T}}=I $ . From this, we can write,
 $ \begin{align}
  & {{A}^{T}}=\dfrac{I}{A} \\
 & {{A}^{T}}=I{{A}^{-1}} \\
 & {{A}^{T}}={{A}^{-1}} \\
\end{align} $
We also know that the determinant of an orthogonal matrix must be equal to $ \pm 1 $ . Here, we have $ \left| A \right|=\pm 1 $ . So, option (a) is true.
Now, we have been given that B is a skew symmetric matrix. Symmetric matrix and a skew symmetric matrix are both square matrices, but the main difference is that, symmetric matrix is equal to its transpose, that is, $ B={{B}^{T}} $ but skew symmetric matrix is a matrix whose transpose is equal to the negative of that matrix, that is, $ {{B}^{T}}=-B $ .
Now, we have \[\left| B \right|=\left| {{B}^{T}} \right|\], as a skew symmetric matrix is also a symmetric matrix. Again, we know that B is a skew symmetric matrix. So, $ {{B}^{T}} $ must be equal to $ \left| -B \right| $ . So, we can write,
\[\begin{align}
  & \left| B \right|=\left| -B \right| \\
 & \left| B \right|=-\left| B \right| \\
\end{align}\]
On transposing \[-\left| B \right|\] from RHS to LHS, we get,
\[\begin{align}
  & \left| B \right|+\left| B \right|=0 \\
 & 2\left| B \right|=0 \\
 & \left| B \right|=0 \\
\end{align}\]
So, option (b) is also true.
Now, we have $ \left| A \right|=\pm 1 $ and \[\left| B \right|=0\]. So, we can calculate \[\left| AB \right|\] as,
\[\left| AB \right|=\left| A \right|\left| B \right|\]
On putting $ \left| A \right|=\pm 1 $ and \[\left| B \right|=0\], we get,
\[\begin{align}
  & \left| AB \right|=\pm 1\times 0 \\
 & \left| AB \right|=0 \\
\end{align}\]
So, we get that option (d) is true and option (c) is false.
So, the false option is “Option c”.

Note: The most common mistake that can occur while solving this question is that, the students get confused between the properties of orthogonal and skew symmetric matrices and may write $ B{{B}^{T}}=I $ for skew symmetric matrix and $ {{A}^{T}}=-A $ for orthogonal matrix. So, the students must be careful while solving this question.