Problem
If M = \( \begin{pmatrix} 5/2 & 3/2 \\ -3/2 & -1/2 \\ \end{pmatrix} \), then which of the following matrices is equal to M2022?
A \( \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \\ \end{pmatrix} \)
B \( \begin{pmatrix} 3034 & -3033 \\ 3033 & -3032 \\ \end{pmatrix} \)
C \( \begin{pmatrix} 3033 & 3032 \\ -3032 & -3031 \\ \end{pmatrix} \)
D \( \begin{pmatrix} 3032 & 3031 \\ -3031 & -3030 \\ \end{pmatrix} \)
Solution
The only hope here is to find a pattern in the powers of M. We’ll cover two methods.
Method 1
Let’s start by evaluating M2, M3, M4, … and see if we get some pattern. I’ll skip the calculations.
M = \( \begin{pmatrix} 5/2 & 3/2 \\ -3/2 & -1/2 \\ \end{pmatrix} \)
M2 = \( \begin{pmatrix} 4 & 3 \\ -3 & -2 \\ \end{pmatrix} \)
M3 = \( \begin{pmatrix} 11/2 & 9/2 \\ -9/2 & -7/2 \\ \end{pmatrix} \)
M4 = \( \begin{pmatrix} 7 & 6 \\ -6 & -5 \\ \end{pmatrix} \)
Do you see it?
The terms in first row and column the powers of M form an AP. And that’s the pattern we were looking for!
The first term of that AP is 5/2 and the common difference is 3/2.
So, the 2022th term would be 5/2 + (2022 – 1) x 3/2 or 3034.
This eliminates options C and D. But that’s not enough – some more work to be done here.
Note that in each power of M, a12 = a11 – 1. That means, option A is the correct one.
We could stop here. But for the sake of completion, let’s verify the pattern for all terms in the matrix.
In each power of M, a21 = -a12 and a22 = a21 + 1. Since the elements of \( \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \\ \end{pmatrix} \) satisfy the relations, option A is true.
Method 2
If we express M = I + A, then it’s easy enough to find higher powers of M, if either of the following relations hold:
- An = A, i.e. A is a periodic matrix
- A2 = A, i.e. A is an idempotent matrix
- An = O, i.e. A is a nilpotent matrix
This is because the binomial theorem is applicable when expanding (A + I)n.
That is, Mn = (I + A)n = nC0I + nC1A + nC2A2 + … nCnAn.
Here, the higher powers of A get simplified using either of the above relations. Let’s give it a try!
We can write M as I + A, where A = \( \begin{pmatrix} 3/2 & 3/2 \\ -3/2 & -3/2 \\ \end{pmatrix} \) or 3/2\( \begin{pmatrix} 1 & 1 \\ -1 & -1 \\ \end{pmatrix} \)
Luckily, A2 comes out to be O. And, using the above expansion, the value of Mn simplifies to I + nA.
So, M2022 equals I + 2022A or \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} + 2022\begin{pmatrix} 3/2 & 3/2 \\ -3/2 & -3/2 \\ \end{pmatrix} \).
This simplifies to \( \begin{pmatrix} 1 + 3033 & 0 + 3033 \\ 0 – 3033 & 1 – 3033 \\ \end{pmatrix}\) or \( \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \\ \end{pmatrix} \).
Therefore, option A is true.
Comments
A nice, moderate problem on matrices – though similar problems have appeared before. If you’re someone who has practiced previous years’ problems, then you wouldn’t find this one too difficult.
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