Let A be a square matrix such that $ {A^T}A = I $ write the value $ \left| A \right| $

Hint: Matrix is the set of the arranged numbers. We get the transpose of the square matrix by changing the rows of the original matrix to column and columns are changed to rows. It is denoted by $ {A^T} $ . The determinant of the matrix is a special number which can be calculated from the square matrix.

Complete step-by-step answer:
Given that - $ {A^T}A = I $
Take determinate on both the sides of the above equation.
Therefore, $ \left| {{A^T}A} \right| = \left| I \right| $
 Determinant of an Identity matrix is always one, irrespective of its order two, three or any.
 $ \therefore \left| {{A^T}} \right| \cdot \left| A \right| = 1 $
 Also, determinant of A and the determinant of the transpose of A is always same or equal, $ \therefore \left| {{A^T}} \right| = \left| A \right| $
 $ \therefore \left| A \right| \cdot \left| A \right| = 1 $
 Product of the two same numbers gives the square.
 $ \therefore {\left| A \right|^2} = 1 $
 Now take square-root on both the sides of the equation
 $ \therefore \sqrt {{{\left| A \right|}^2}} = \sqrt 1 $
Square and square-root cancels each other on the left hand side of the equation and on the right hand side of the equation the square root of the number one gives us the value of plus or minus one.
 $ \therefore \left| A \right| = \pm 1 $
Therefore, the required answer - When A be a square matrix such that $ {A^T}A = I $ then the value $ \left| A \right| = \pm 1 $

Note: Simply, the transpose of the matrix is the flipped version of the original matrix. It was introduced by the British Mathematician Arthur Cayley in the year 1858. Also, since the value of the numbers in the matrix remains the same, the determinant of the matrix and its transpose remains the same. Determinant can also be denoted by $\det (A),{\text{ det A or }}\left| A \right|$