Q80 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { If the matrix } A=\left(\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right) \text { then } A^{n+1}=$
- Correct answerA$2^n\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$
- B$2 n\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$
- C$2\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$
- D$2^{n+1}\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$
Solution
Given $A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$.
Calculate $A^2$:
$A^2 = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix} = 2 \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = 2A$.
Calculate $A^3$:
$A^3 = A^2 \cdot A = (2A) \cdot A = 2A^2 = 2(2A) = 2^2 A$.
By induction, assume $A^k = 2^{k-1} A$.
Then $A^{k+1} = A^k \cdot A = (2^{k-1} A) \cdot A = 2^{k-1} A^2 = 2^{k-1} (2A) = 2^k A$.
Thus, $A^n = 2^{n-1} A$ for $n \ge 1$.
Substituting $n+1$ for $n$, we get $A^{n+1} = 2^{(n+1)-1} A = 2^n A$.
Therefore, $A^{n+1} = 2^n \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$.
Answer: $2^n\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$
Calculate $A^2$:
$A^2 = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix} = 2 \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = 2A$.
Calculate $A^3$:
$A^3 = A^2 \cdot A = (2A) \cdot A = 2A^2 = 2(2A) = 2^2 A$.
By induction, assume $A^k = 2^{k-1} A$.
Then $A^{k+1} = A^k \cdot A = (2^{k-1} A) \cdot A = 2^{k-1} A^2 = 2^{k-1} (2A) = 2^k A$.
Thus, $A^n = 2^{n-1} A$ for $n \ge 1$.
Substituting $n+1$ for $n$, we get $A^{n+1} = 2^{(n+1)-1} A = 2^n A$.
Therefore, $A^{n+1} = 2^n \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$.
Answer: $2^n\left(\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right)$
Hence, the correct answer is option: 'A'.