Question

Let $A = \begin{bmatrix} 1 & 2 \\\ 0 & 1 \end{bmatrix}$ and $B = I + \text{adj}(A) + (\text{adj}(A))^2 + \dots + (\text{adj}(A))^{10}.$Then, the sum of all the elements of the matrix $B$ is:

• A. -110
• B. 22
• C. -88
• D. -124

Answer 1

Answer: (C)

-88

Answer 2

We are given that $A = \begin{pmatrix} 1 & 2 \\\ 0 & 1 \end{pmatrix}$. The adjugate matrix $\text{adj}(A)$ is defined as the transpose of the cofactor matrix of $A$.

First, calculate $\text{adj}(A)$:

$\text{adj}(A) = \begin{pmatrix} 1 & -2 \\\ 0 & 1 \end{pmatrix}$

Next, we calculate $\text{adj}(A)^2$:

$\text{adj}(A)^2 = \begin{pmatrix} 1 & -2 \\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\\ 0 & 1 \end{pmatrix}$

Then, we calculate $\text{adj}(A)^{10}$, which follows a similar process:

$\text{adj}(A)^{10} = \begin{pmatrix} 1 & -20 \\\ 0 & 1 \end{pmatrix}$

The matrix $B$ is the sum of these matrices:

$B = I + \text{adj}(A) + \text{adj}(A)^2 + \cdots + \text{adj}(A)^{10}$

$B = \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & -2 \\\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & -4 \\\ 0 & 1 \end{pmatrix} + \cdots + \begin{pmatrix} 1 & -20 \\\ 0 & 1 \end{pmatrix}$

Summing the elements of $B$, we find:

$B = \begin{pmatrix} 11 & -110 \\\ 0 & 11 \end{pmatrix}$

Thus, the sum of all elements of $B$ is:

$11 + (-110) + 11 = -88$