← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

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

Question

If $A=\begin{bmatrix}1 & -1 & 1\\ 2 & -1 & 0\\ 1 & 0 & 0\end{bmatrix}$, show that $A^2=A^{-1}$ without finding $A^{-1}$.

Technique

$A^2=A^{-1}\Leftrightarrow A^3=I$; compute $A^3$ directly.

Solution

Strategy. $A^2=A^{-1}$ iff $A^3=I$. So compute $A^3$ and check.

Step 1 — Compute $A^2$

$A^2=A\cdot A$. Row by row:

Row 1 of $A^2$: $(1)(1,2,1)+(-1)(-1,-1,0)+(1)(1,0,0)$… wait, easier to use column form.

$A=\begin{bmatrix}1 & -1 & 1\\ 2 & -1 & 0\\ 1 & 0 & 0\end{bmatrix}$.

$(A^2)_{ij}=\sum_k A_{ik}A_{kj}$.

Row 1 of $A^2$:

• $(A^2)_{11}=1\cdot 1+(-1)\cdot 2+1\cdot 1=1-2+1=0$.
• $(A^2)_{12}=1\cdot(-1)+(-1)(-1)+1\cdot 0=-1+1+0=0$.
• $(A^2)_{13}=1\cdot 1+(-1)\cdot 0+1\cdot 0=1$.

Row 2 of $A^2$:

• $(A^2)_{21}=2\cdot 1+(-1)\cdot 2+0\cdot 1=0$.
• $(A^2)_{22}=2\cdot(-1)+(-1)(-1)+0\cdot 0=-1$.
• $(A^2)_{23}=2\cdot 1+(-1)\cdot 0+0\cdot 0=2$.

Row 3 of $A^2$:

• $(A^2)_{31}=1\cdot 1+0\cdot 2+0\cdot 1=1$.
• $(A^2)_{32}=1\cdot(-1)+0\cdot(-1)+0\cdot 0=-1$.
• $(A^2)_{33}=1\cdot 1+0\cdot 0+0\cdot 0=1$.

Step 2 — Compute $A^3=A\cdot A^2$

Row 1: $1\cdot(0,0,1)+(-1)(0,-1,2)+1\cdot(1,-1,1)=(0,0,1)+(0,1,-2)+(1,-1,1)=(1,0,0)$.

Row 2: $2\cdot(0,0,1)+(-1)(0,-1,2)+0\cdot(1,-1,1)=(0,0,2)+(0,1,-2)+(0,0,0)=(0,1,0)$.

Row 3: $1\cdot(0,0,1)+0\cdot(0,-1,2)+0\cdot(1,-1,1)=(0,0,1)$.

Step 3 — Conclude

$A^3=I\Rightarrow A\cdot A^2=I\Rightarrow A^2=A^{-1}$. $\blacksquare$

Answer