Question

Find the transpose($A^T$) of the matrix A, if \[A = \left( {\begin{array}{*{20}{c}}
  0&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)\].
A) \[A\]
B) \[ - A\]
C) \[I\]
D) \[{A^2}\]

Answer 1

Hint: In order to find \[{A^T}\] for the given matrix \[A\], first we should know what exactly \[{A^T}\] is. The \[{A^T}\] represents Transpose of a matrix. Transpose of a matrix is also a matrix in which the rows and columns of the original matrix are exchanged or swapped.

Complete step by step solution:
We are given with a \[3 \times 3\]matrix \[A = \left( {\begin{array}{*{20}{c}}
  0&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)\], where \[3 \times 3\] represents the number of rows and columns.

Since, we know that Transpose of a matrix means flipping the matrix over its diagonal. Means the rows and column values are swapped, rows become columns and columns become rows. For example: - The 3rd value of the first row becomes the 3rd value of the first column and the 3rd value of the first column becomes the 3rd value of the first row. And, similarly this happens for the other rows and columns.
So, according to the definition, to find the transpose of the matrix \[A\], we are swapping the 1st whole row with the first column, all the values of the second row with second column and the values of third row with third column and we get:
\[
  A = \left( {\begin{array}{*{20}{c}}
  0&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right) \\
   \Rightarrow {A^T} = \left( {\begin{array}{*{20}{c}}
  0&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right) \\
 \]
Which is our Transpose \[{A^T}\] for the matrix \[A\].

But we can see that the values of the matrix and the values of the transpose of the matrix are quite similar, so making a relation between them.
For that multiplying and dividing all the values of the transpose with \[ - 1\]:
\[
  {A^T} = \left( {\begin{array}{*{20}{c}}
  0&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right) \\
   \Rightarrow {A^T} = \dfrac{{ - 1}}{{ - 1}}\left( {\begin{array}{*{20}{c}}
  0&{ - 1}&{ - 4} \\
  1&0&{ - 7} \\
  4&7&0
\end{array}} \right) \\
 \]
Further solving the matrix by taking \[\dfrac{1}{{ - 1}}\] inside the matrix and multiplying each term of the matrix with this value and we get:
\[{A^T} = - 1\left( {\begin{array}{*{20}{c}}
  {\dfrac{0}{{ - 1}}}&{\dfrac{{ - 1}}{{ - 1}}}&{\dfrac{{ - 4}}{{ - 1}}} \\
  {\dfrac{1}{{ - 1}}}&{\dfrac{0}{{ - 1}}}&{\dfrac{{ - 7}}{{ - 1}}} \\
  {\dfrac{4}{{ - 1}}}&{\dfrac{7}{{ - 1}}}&{\dfrac{0}{{ - 1}}}
\end{array}} \right)\]
On further solving we get:
\[{A^T} = - 1\left( {\begin{array}{*{20}{c}}
  0&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)\]
And, we can see that the value inside the matrix box above is equal to \[A\], that means the relation between the transpose and the matrix is:
\[
  {A^T} = - 1\left( A \right) \\
   \Rightarrow {A^T} = - A \\
 \]
Which matches with our second option.

Therefore, \[A = \left( {\begin{array}{*{20}{c}}
  0&1&4 \\
  { - 1}&0&7 \\
  { - 4}&{ - 7}&0
\end{array}} \right)\], then \[{A^T} = - A\].

Therefore, the given matrix satisfies \[{A^T} = - A\]. So, option (B) is correct.

Note:
> When \[{A^T} = - A\], then Matrix A is called a skew symmetric matrix.
> Transpose of a matrix \[{A^T}\] is also represented as \[A'\].
> With the same methods we can find the transpose of any matrix given.
> There is no such confirmation, that there will always be a relation between matrix and its transpose.