Let [A = ( pmatrix 1 & 0 & 0 0 & 4 & - 1 0 & 12 & - 3 pmatrix )].

Question

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

Accepted Answer

$A^{2}=\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{array}\right]$

$$
=\left[\[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 4 & -1 \ 0 & 12 & -3 \end{array}$$\right]=A
\]

$$\Rightarrow \mathrm{A}_{3}=\mathrm{A}_{4}=.......=\mathrm{A}$$

Now,

$$
\begin{aligned}
(A+I)^{11} & ={ }^{11} C_{0} A^{11}+{ }^{11} C_{1} A^{10}+\ldots{ }^{11} C_{11} I \\
& =A\left({ }^{11} C_{0}+{ }^{11} C_{1} \ldots{ }^{11} C_{10}\right)+I \\
& =A\left(2^{11}-1\right)+I
\end{aligned}
$$

Trace of

$$
\begin{aligned}
(A+I)^{11} & =2^{11}+4\left(2^{11}-1\right)+1-3\left(2^{11}-1\right)+1 \\
& =2 \times 2^{11}-4+3+2 \\
& =2^{12}+1 \\
& =4097
\end{aligned}
$$

Solutions · 1