UPSC 2019 Maths Optional Paper 1 Q2b — Step-by-Step Solution

15 marks · Section A

Orthogonal and unitary matrices · Linear Algebra · asked 3× in 13 yrs · Read the full method →

Question

Let $A$ and $B$ be two orthogonal matrices of same order and $\det A+\det B=0$ . Show that $A+B$ is a singular matrix.

Technique

Orthogonality $\Rightarrow\det=\pm1$ ; opposite signs $\Rightarrow\det A\det B=-1$ ; the identity $A+B=A(A+B)^TB$ then forces $\det(A+B)=-\det(A+B)=0$ .

Solution

Let $A,B$ be $n\times n$ orthogonal matrices: $AA^T=BB^T=I$ , so $\det A,\det B\in\{+1,-1\}$ . The hypothesis $\det A+\det B=0$ forces them to have opposite determinants:

\det A=-\det B,\qquad\text{i.e.}\qquad \det A\cdot\det B=-1.

Step 1 — The key factorisation

Write, using $B B^T=I$ (so $B^T=B^{-1}$ ) and $A A^T=I$ :

A+B = A(B^T+A^T)B = A\,(A+B)^T\,B. \tag{$\ast$}

Indeed $A(B^T+A^T)B = AB^TB+AA^TB = A\cdot I + I\cdot B = A+B$ , since $B^TB=I$ and $AA^T=I$ . ✓ And $B^T+A^T=(A+B)^T$ .

Step 2 — Take determinants

From $(\ast)$ ,

\det(A+B)=\det A\,\cdot\,\det\!\big((A+B)^T\big)\,\cdot\,\det B=\det A\,\det B\,\det(A+B),

because $\det\big((A+B)^T\big)=\det(A+B)$ . Using $\det A\,\det B=-1$ :

\det(A+B)=-\det(A+B)\ \Longrightarrow\ 2\det(A+B)=0\ \Longrightarrow\ \det(A+B)=0.

Answer

\boxed{\;\det(A+B)=0,\ \text{so } A+B \text{ is singular.}\;}