Question

Let $A=\begin{pmatrix}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{pmatrix}$ Then the sum of the diagonal elements of the matrix $(A+I)^{11}$ is equal to :

• A. 2050
• B. 4094
• C. 6144
• D. 4097

Hint:

For matrix power problems, use the binomial expansion for matrices and simplify using powers of the matrix \( A \).

Answer 1

The Correct Option is D

$A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& -1\\ 0& 12& -3\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& -1\\ 0& 12& -3\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 4& -1\\ 0& 12& -3\end{array}\right]=A$
$\Rightarrow A^3=A^4=\dots \dots =A$
$(A+I){}^{11}={}^{11}{C}_0{A}^{11}+{}^{11}{C}_1{A}^{10}+\dots .\cdot {}^{11}{C}_{10}A+{}^{11}{C}_{11}I$
$=\left({}^{11}{C}_0+{}^{11}{C}_1+\dots .{}^{11}{C}_{10}\right)A+I$
$=\left(2^{11}-1\right)A+I=2047A+I$
$\therefore$ Sum of diagonal elements $=2047(1+4-3)+3$
$=4094+3=4097$

Answer 2

Step 1: First, calculate \( A^2 \): \[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix} \] This matrix multiplication yields the same matrix \( A \), meaning that \( A^2 = A \). 
Step 2: From this, we have: \[ A^3 = A^4 = \cdots = A \] Since \( A^2 = A \), any higher powers of \( A \) will also be equal to \( A \). 
Step 3: Now, calculate \( (A + I)^{11} \): \[ (A + I)^{11} = \sum_{k=0}^{11} \binom{11}{k} A^k I^{11-k} = \left( (2^{11} - 1)A + I \right) \] Here, we used the binomial expansion of \( (A + I)^{11} \), and since \( A^k = A \) for all \( k \geq 1 \), the expression simplifies as shown. 
Step 4: The sum of the diagonal elements is then: \[ \text{Sum of diagonal elements} = 2047 \times (1 + 4 - 3) + 3 = 4094 + 3 = 4097 \] This gives the final result for the sum of the diagonal elements of \( (A + I)^{11} \).