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 :
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 :
Show Hint
- 2050
- 4094
- 6144
- 4097
The Correct Option is D
Approach Solution - 1
Sum of diagonal elements
Approach Solution -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 \( (A + I)^{11} \).
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Concepts Used:
Matrix Transformation
The numbers or functions that are kept in a matrix are termed the elements or the entries of the matrix.
Transpose Matrix:
The matrix acquired by interchanging the rows and columns of the parent matrix is termed the Transpose matrix. The definition of a transpose matrix goes as follows - “A Matrix which is devised by turning all the rows of a given matrix into columns and vice-versa.”