Question:In Exercises 31–36, mention an appropriate theorem in your explanation.

36. Let U be a square matrix such that \({U^T}U = I\). Show that\(det{\rm{ }}U = \pm 1\).

According to theorem 6, if A andB aresquare matrices, then thedeterminant of the product matrix AB is equal to the product of the determinant of A and the determinant of B.

\(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)

If matrices A and B are the same, then the general form is \(\det {A^n} = {\left( {\det A} \right)^n}\).

According to theorem 5, the determinant of square matrix A is equal to the determinant of the transpose matrix A.

\(\det {A^T} = \det A\)

Apply the theorem \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\) by substituting \(A = {U^T}\)and \(B = U\), as shown below:

\(\det \left( {{U^T}U} \right) = \left( {\det {U^T}} \right)\left( {\det U} \right)\)

Apply the theorem \(\det {A^T} = \det A\).

\(\begin{aligned}{}\det \left( {{U^T}U} \right) &= \left( {\det U} \right)\left( {\det U} \right)\\ &= {\left( {\det U} \right)^2}\end{aligned}\)

It is given that \({U^T}U = I\). By using this,

\(\begin{aligned}{}\det \left( {{U^T}U} \right) &= {\left( {\det U} \right)^2}\\{\left( {\det U} \right)^2} &= \det \left( I \right)\\{\left( {\det U} \right)^2} &= 1\\\det U &= \pm 1.\end{aligned}\)

Hence, it is proved that \(\det U = \pm 1\).