Question

If $A = \left( {\begin{array}{*{20}{c}}
  x&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)$ such that ${A^T} = - A$ ,then $x = $
A. $ - 1$
B. $0$
C. $1$
D. $4$

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve this problem.
Here, we are given a square matrix. And, we are asked to calculate the value of $x$.
We can call a matrix a square matrix whose rows and columns are equal.
Also, we are given a condition to obtain the answer. We need to compare the transpose matrix and the square matrix $ - A$.

Complete step by step answer:
It is given that $A = \left( {\begin{array}{*{20}{c}}
  x&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)$
We are asked to calculate the value of $x$ provided ${A^T} = - A$
Now, we shall calculate the transpose of $A$ .
${A^T} = {\left( {\begin{array}{*{20}{c}}
  x&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)^T}$
${A^T} = \left( {\begin{array}{*{20}{c}}
  x&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right)$
Also, we are given a condition to ${A^T} = - A$ solve this problem.
Now, we shall find the matrix $ - A$
$ - A = - \left( {\begin{array}{*{20}{c}}
  x&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)$
$ \Rightarrow - A = \left( {\begin{array}{*{20}{c}}
  { - x}&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right)$
Now, we shall apply the obtained matrices in the given condition.
${A^T} = - A$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  x&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - x}&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right)$
We can note that in the above equation, all the values in the rows and columns are equal.
So, we can compare the value of $x$
Thus, we get
$ \Rightarrow x = - x$
$ \Rightarrow x + x = 0$
$ \Rightarrow 2x = 0$
$ \Rightarrow x = 0$
Hence, the value of $x$ is zero.

So, the correct answer is “Option B”.

Note:
 The transpose of a matrix can be applied only if the given matrix is square or else the order of the given matrix will be changed. Since the matrix given in this problem is a square matrix, we are able to find the transpose of a matrix.
Also, if the values in the rows and columns are equal in the matrix equation, we are able to compare the unknown values. Hence, the value of $x$ is zero.