IPMAT Indore 2022 · Quantitative Ability · Question 26
IPMAT Indore 2022 previous year question on Matrices and Determinants, with the answer and a step-by-step solution.
- ANone of these
- B\left[ \begin{matrix}1 & 0 \\ 1011 & 0\end{matrix}\right]
- C\left[ \begin{matrix}1 & 0 \\ 2022 & 0\end{matrix}\right]
- D\left[ \begin{matrix}1 & 0 \\ \frac{1}{{2}^{2022}}& 0\end{matrix}\right]
Answer & solution
Answer: Option B
This question is defective and was discarded. The working below shows why, and what was almost certainly intended.
Step 1 — compute {A}^{2} from the matrix as printed.
{A}^{2}=\left[ \begin{matrix}1 & 0 \\ 1/2 & 0\end{matrix}\right]\left[ \begin{matrix}1 & 0 \\ 1/2 & 0\end{matrix}\right]=\left[ \begin{matrix}1\cdot 1+0\cdot 1/2 & 0 \\ 1/2\cdot 1+0\cdot 1/2 & 0\end{matrix}\right]=\left[ \begin{matrix}1 & 0 \\ 1/2 & 0\end{matrix}\right]=A
So A is idempotent: {A}^{2}=A, and therefore
{A}^{n}=A \text{for every}n\ge 1 \Rightarrow{A}^{2022}=\left[ \begin{matrix}1 & 0 \\ 1/2 & 0\end{matrix}\right]
Step 2 — compare with the options. That matrix is not option 2, 3 or 4. Taken exactly as printed, the only defensible answer is Option 1, “None of these”.
Step 3 — what was intended. Replace the bottom-right 0 with 1:
B=\left[ \begin{matrix}1 & 0 \\ 1/2 & 1\end{matrix}\right] \Rightarrow{B}^{n}=\left[ \begin{matrix}1 & 0 \\ n/2 & 1\end{matrix}\right]
(a shear, proved in one line by induction). Then {B}^{2022}=\left[ \begin{matrix}1 & 0 \\ 1011 & 1\end{matrix}\right] — and 1011=2022/2 is exactly the entry in option 2, which carries a 0 where it should carry a 1. So the misprint is in both the question and the key.
Answer: Option 2 is the published answer and the one that scores. Mathematically, the printed matrix gives “None of these”.
Shortcut. Treat this as a free mark and move on. It appears in the published keys as option 2, so mark that; but note for teaching that the printed matrix satisfies {A}^{2}=A, which makes every power equal to A itself.
The genuine skill being tested is worth keeping, because the intended version is a standard result:
{\left[ \begin{matrix}1 & 0 \\ k & 1\end{matrix}\right]}^{n}=\left[ \begin{matrix}1 & 0 \\ nk & 1\end{matrix}\right]
A unipotent 2\times 2 shear simply multiplies its off-diagonal entry by n. With k=1/2 and n=2022 that gives 1011.
And the broader habit: always compute {A}^{2} first. It instantly tells you whether a matrix is idempotent ({A}^{2}=A), involutory ({A}^{2}=I) or nilpotent ({A}^{2}=0) — each of which collapses a huge power to something trivial.
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