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

Answer 1

Answer: (D)

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