Matrices & Determinants
Orthogonal Matrix — Algebraic Identity
nta_pyq_2024_jan
Grade 12
Question:
Let $A$ be a square matrix such that $AA^T=I$. Then $\dfrac{1}{2}A\left[(A+A^T)^2+(A-A^T)^2\right]$ is equal to
$A^2+I$
$A^3+I$
$A^2+A^T$
$A^3+A^T$
Step-by-Step Solution
Key Concept: Expand $(A+A^T)^2+(A-A^T)^2=2A^2+2(A^T)^2$. Then $\frac{1}{2}A[2A^2+2(A^T)^2]=A[A^2+(A^T)^2]=A^3+A(A^T)^2$. Use $AA^T=I\Rightarrow A(A^T)^2=AA^T\cdot A^T=I\cdot A^T=A^T$.
$\frac{1}{2}A[(A+A^T)^2+(A-A^T)^2]=A[A^2+(A^T)^2]=A^3+A(A^T)^2=A^3+A^T$.
Correct Answer: 4