IPMAT Indore 2022 · Quantitative Ability · Question 7

Step 1 — see what A does. A is a permutation matrix: it fixes the first coordinate and swaps the second and third. Swapping twice returns you to the start, so

{A}^{2}=I

Step 2 — reduce every power. Because {A}^{2}=I, powers alternate:

{A}^{9}={A}^{8}\cdot A=A,{A}^{6}=I,{A}^{3}={A}^{2}\cdot A=A

Step 3 — add them up.

{A}^{9}+{A}^{6}+{A}^{3}+A=A+I+A+A=3A+I

Step 4 — write out the matrix.

3A+I=\left[ \begin{matrix}3 & 0 & 0 \\ 0 & 0 & 3 \\ 0 & 3 & 0\end{matrix}\right]+\left[ \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right]=\left[ \begin{matrix}4 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 3 & 1\end{matrix}\right]

Step 5 — take the determinant. Expanding along the first row:

\det =4\left( 1\cdot 1-3\cdot 3 \right)=4\left( 1-9 \right)=-32

\left| \det \right|=32

Answer: 32

Shortcut. Check {A}^{2} first — always. For a permutation matrix built from a single swap, {A}^{2}=I, and the whole sum collapses to 3A+I in one line.

Then exploit the block structure: the matrix is 4 in the top-left corner and the 2\times 2 block \left[ \begin{matrix}1 & 3 \\ 3 & 1\end{matrix}\right] below it, so

\det =4\left({1}^{2}-{3}^{2}\right)=-32

A matrix that is its own inverse turns any polynomial in A into just \alpha A+\beta I — worth recognising, because it reduces a ninth power to arithmetic.