Question

If $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$ and $A^{10} = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$ then

$\Sigma a_1 + \Sigma b_1 + \Sigma c_1$ equals

• A. 11
• B. 22
• C. 33
• D. 44

Answer 1

Answer: (C)

33

Answer 2

The given matrix is $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$. We want to find $A^{10}$.

Let's write $A$ as the sum of the identity matrix $I$ and a matrix $N$: $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} = I + N$.

Let's compute the powers of $N$: $N^1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix}$ $N^2 = N \cdot N = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

Since $N^2 = \mathbf{0}$, $N$ is a nilpotent matrix of index 2. We want to compute $A^{10} = (I + N)^{10}$. Since $I$ commutes with any matrix (including $N$), we can use the binomial theorem for matrices:

$(I + N)^{10} = \binom{10}{0} I^{10} N^0 + \binom{10}{1} I^9 N^1 + \binom{10}{2} I^8 N^2 + \dots + \binom{10}{10} I^0 N^{10}$.

Since $N^2 = \mathbf{0}$, all terms with $N^k$ for $k \ge 2$ are zero matrices. So, $(I + N)^{10} = \binom{10}{0} I + \binom{10}{1} N$. $\binom{10}{0} = 1$ and $\binom{10}{1} = 10$. $A^{10} = 1 \cdot I + 10 \cdot N = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + 10 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 10 & 20 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 10 & 20 & 1 \end{bmatrix}$.

The matrix $A^{10}$ is given as $\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$. So, $a_1 = 1, b_1 = 0, c_1 = 0$. $a_2 = 0, b_2 = 1, c_2 = 0$. $a_3 = 10, b_3 = 20, c_3 = 1$.

The expression to evaluate is $\Sigma a_1 + \Sigma b_1 + \Sigma c_1$. The notation $\Sigma a_1$ likely refers to the sum of elements in the first column, $\Sigma b_1$ to the sum of elements in the second column, and $\Sigma c_1$ to the sum of elements in the third column.

Sum of elements in the first column = $a_1 + a_2 + a_3 = 1 + 0 + 10 = 11$. Sum of elements in the second column = $b_1 + b_2 + b_3 = 0 + 1 + 20 = 21$. Sum of elements in the third column = $c_1 + c_2 + c_3 = 0 + 0 + 1 = 1$. The required sum is $(a_1+a_2+a_3) + (b_1+b_2+b_3) + (c_1+c_2+c_3) = 11 + 21 + 1 = 33$.