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UPSC 2016 Maths Optional Paper 1 Q1a-ii — Step-by-Step Solution

4 marks · Section A

Algebra of matrices · Linear Algebra · asked 5× in 13 yrs · Read the full method →

Question

If $A=\begin{bmatrix}1 & 1 & 3\\ 5 & 2 & 6\\ -2 & -1 & -3\end{bmatrix}$, then find $A^{14}+3A-2I$.

Technique

Recognize nilpotency by computing successive powers; $A^3=O$ collapses $A^{14}$ to $O$, leaving the linear part $3A-2I$.

Solution

Step 1 — Test for nilpotency

Compute $A^2$:

Not yet zero. Compute $A^3=A\cdot A^2$:

So $A$ is nilpotent of index 3: $A^3=O$.

Step 2 — Reduce $A^{14}$

Since $A^3=O$, every power $A^k=O$ for $k\ge3$. In particular $A^{14}=A^{12}\cdot A^2=(A^3)^4\,A^2=O$.

Step 3 — Evaluate the expression

Answer