Let A=[0 -2,2 0], M=ΣA²ᵏ, N=ΣA²ᵏ⁻¹ (k=1 to 10): Find MN²

JEE Main 2022 — 25 June, Shift 1. Previous Year Question.

Problem

Let $A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$. If $M$ and $N$ are two matrices given by $M = \displaystyle\sum_{k=1}^{10} A^{2k}$ and $N = \displaystyle\sum_{k=1}^{10} A^{2k-1}$, then $MN^2$ is:

Key insight. Computing ten separate powers of a matrix sounds painful, but $A^2$ turns out to be a plain number times the identity matrix — and once that’s known, every higher power of $A$ collapses into “that same number, raised to a power, times either $A$ or the identity.” The whole problem becomes a geometric series in ordinary numbers.

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Approach

The standing question with any matrix-power series is: does a pattern emerge after computing just the first couple of powers? Here, squaring $A$ gives a multiple of the identity matrix $I$, which is the best possible outcome — it means $A^3 = A^2 \cdot A$ is just a number times $A$, $A^4 = A^2\cdot A^2$ is a number times $I$ again, and so on, alternating between “number times $A$” and “number times $I$” forever. That turns both $M$ (even powers) and $N$ (odd powers) into geometric series that can be summed with the standard formula, entirely avoiding repeated matrix multiplication.

Solution

Step 1 — Compute $A^2$

$$A^2 = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} -4 & 0 \\ 0 & -4 \end{pmatrix} = -4I$$

This is the key fact: $A^2 = -4I$.

Step 2 — Find the pattern for higher powers

$$A^3 = A^2\cdot A = -4I\cdot A = -4A, \qquad A^4 = A^2\cdot A^2 = (-4I)(-4I) = 16I = (-4)^2 I$$

In general, $A^{2k} = (-4)^k I$ and $A^{2k-1} = (-4)^{k-1}A$.

Step 3 — Sum the series for $M$

$$M = \sum_{k=1}^{10} A^{2k} = \sum_{k=1}^{10} (-4)^k I = I\sum_{k=1}^{10}(-4)^k$$

This is a geometric series with first term $-4$, common ratio $-4$, and $10$ terms:

$$\sum_{k=1}^{10}(-4)^k = \frac{-4\big((-4)^{10}-1\big)}{-4-1} = \frac{(-4)^{10}-1}{5}\cdot(-4)\Big/(-1)$$

Simplified directly using $a\dfrac{r^n-1}{r-1}$ with $a=-4$, $r=-4$, $n=10$: since $(-4)^{10} = 4^{10}$ (even power, so positive),

$$M = \frac{4^{10}-1}{5}\, I$$

Step 4 — Sum the series for $N$

$$N = \sum_{k=1}^{10} A^{2k-1} = \sum_{k=1}^{10} (-4)^{k-1}A = A\sum_{k=0}^{9}(-4)^{k}$$

This geometric series has first term $1$, ratio $-4$, $10$ terms:

$$\sum_{k=0}^{9}(-4)^k = \frac{(-4)^{10}-1}{-4-1} = \frac{4^{10}-1}{-5} = -\frac{4^{10}-1}{5}$$

So $N = -\dfrac{4^{10}-1}{5}A$.

Step 5 — Compute $N^2$

Let $c = \dfrac{4^{10}-1}{5}$ for brevity, so $N = -cA$. Then:

$$N^2 = c^2 A^2 = c^2(-4I) = -4c^2 I$$

Step 6 — Compute $MN^2$

$$MN^2 = (cI)(-4c^2 I) = -4c^3 I$$

Writing this out with $I = \begin{pmatrix}1&0\\0&1\end{pmatrix}$, the result is a matrix of the form $\begin{pmatrix}k&0\\0&k\end{pmatrix}$ for some constant $k = -4c^3$ — a scalar multiple of the identity, but since $k \ne 1$, it is not itself an identity matrix. It’s a diagonal (in fact, symmetric) matrix, not a skew-symmetric one, since its diagonal entries are equal and non-zero.

Answer

Common mistakes

• Trying to compute $A^2, A^3, \ldots, A^{20}$ one at a time instead of spotting that $A^2=-4I$ collapses the whole problem into a geometric series — technically possible, but enormously slower and error-prone.
• Mixing up which sum is geometric in which variable. $M$ sums $(-4)^k$ directly; $N$ sums $(-4)^{k-1}$, which shifts the starting index — losing track of this off-by-one changes the final constant.

Practise next

• For $B = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, find $B^2$ and use the same approach to sum $\sum_{k=1}^{5} B^{2k}$.