Question

If A is a square matrix satisfying $\text{{A}'A}=\text{I}$ , write the value of $\left| \text{A} \right|$ .

Answer 1

Hint: Firstly, we have to take the determinant on both the sides of the given equation, $\text{{A}'A}=\text{I}$ . Then, we have to apply the properties of determinant mainly $\left| AB \right|=\left| A \right|\left| B \right|$ and $\left| A \right|=\left| {{A}'} \right|$ . We will use the property that the determinant of the identity matrix is always 1. Now, we have to simplify the resultant equation.

Complete step by step solution:
We are given that $\text{{A}'A}=I$ . Let us take determinants on both sides.
$\Rightarrow \left| {A}'A \right|=\left| I \right|$
We know that if A and B are square matrix of same order, then $\left| AB \right|=\left| A \right|\left| B \right|$ . We also know that determinant of identity matrix is always 1, that is, $\left| I \right|=1$ .Therefore, we can write the above equation as
$\Rightarrow \left| {{A}'} \right|\left| A \right|=1$
We know that for any square matrix, A we can write $\left| A \right|=\left| {{A}'} \right|$ . Therefore, the above equation can be written as
$\Rightarrow \left| A \right|\left| A \right|=1$
We can write the LHS as
$\Rightarrow {{\left| A \right|}^{2}}=1$
Let us take square roots on both sides. We can write the result of this step as
$\Rightarrow \left| A \right|=\pm 1$
Therefore, the value of $\left| \text{A} \right|$ is $\pm 1$ .

Note: Students must be thorough with the properties of determinants. We can only apply the property $\left| AB \right|=\left| A \right|\left| B \right|$ only if A and B are square matrices of the same order. Similarly, we can only apply the property $\left| A \right|=\left| {{A}'} \right|$ if A is a square matrix. Students must never miss to put the $\pm $ sign after finding the square root in the last step. We also represent the transpose of a matrix A as ${{A}^{\text{T}}}$ .