Matrices & Determinants
Matrix multiplication
Grade Class 12

Question:

If A = $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ & A$^n$ = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, (where n $\ge$ 2 & n $\in$ N), then -
(A) a = d
(B) b = c
(C) b = a + 1 if n is odd
(D) b = a - 1 if n is even

Step-by-Step Solution

Key Concept: The matrix A is symmetric, so A^n will also be symmetric, implying a=d and b=c. By calculating powers of A, one can observe the pattern for a and b.
Since A = $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ is symmetric, A$^n$ is also symmetric for all n $\in$ N. Thus, a = d and b = c. Calculating A$^2$ = $\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}$, A$^3$ = $\begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}$. For n=2 (even), b = a-1 (4 = 5-1). For n=3 (odd), b = a+1 (14 = 13+1).
Correct Answer: 1,2,3,4

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