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

The Correct Option is C

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:

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

Answer 2

Step 1: Understand the given matrices.
We are given matrix \( A \) as:
\[ A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \] We are also given matrix \( B \) as the sum of powers of the adjugate of \( A \), from \( \text{adj}(A) \) to \( (\text{adj}(A))^{10} \), plus the identity matrix \( I \). That is: \[ B = I + \text{adj}(A) + (\text{adj}(A))^2 + \dots + (\text{adj}(A))^{10} \] We need to find the sum of all the elements of matrix \( B \).

Step 2: Compute the adjugate of \( A \).
The adjugate of a matrix \( A \), denoted \( \text{adj}(A) \), is the transpose of the cofactor matrix. For a 2x2 matrix: \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \quad \text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For matrix \( A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \), the adjugate is: \[ \text{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \] This is because the cofactor matrix of \( A \) is \( \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \), and the adjugate is the transpose of this matrix.

Step 3: Observe the powers of the adjugate matrix.
Now, we examine the powers of \( \text{adj}(A) \): \[ \text{adj}(A) = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \] Let's calculate the first few powers of \( \text{adj}(A) \):
- \( \text{adj}(A)^2 = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix} \) - \( \text{adj}(A)^3 = \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} \) You can observe that each subsequent power of \( \text{adj}(A) \) increases the second element of the top row by -2. Therefore, for \( k \geq 1 \), the general form of \( \text{adj}(A)^k \) is: \[ \text{adj}(A)^k = \begin{bmatrix} 1 & -2k \\ 0 & 1 \end{bmatrix} \] Thus, \( (\text{adj}(A))^k \) has the general form where the second element of the top row is \( -2k \).

Step 4: Compute matrix \( B \).
Now, matrix \( B \) is the sum of \( I \) and powers of \( \text{adj}(A) \): \[ B = I + \text{adj}(A) + (\text{adj}(A))^2 + \dots + (\text{adj}(A))^{10} \] Substituting the values: \[ B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix} + \dots + \begin{bmatrix} 1 & -20 \\ 0 & 1 \end{bmatrix} \] Let's break down the sum row by row:
- The sum of the first row is: \[ 1 + 1 + 1 + \dots + 1 \quad \text{(11 terms)} \quad = 11 \] and \[ 0 + (-2) + (-4) + \dots + (-20) \quad = -2(1 + 2 + 3 + \dots + 10) = -2 \times 55 = -110 \] So, the first row of \( B \) is \( [11, -110] \).
- The sum of the second row is: \[ 0 + 0 + 0 + \dots + 0 \quad = 0 \] and \[ 1 + 1 + 1 + \dots + 1 \quad \text{(11 terms)} \quad = 11 \] So, the second row of \( B \) is \( [0, 11] \).

Step 5: Sum of all elements of \( B \).
The sum of all the elements of matrix \( B \) is: \[ 11 + (-110) + 0 + 11 = -88 \] Thus, the sum of all the elements of matrix \( B \) is \( \boxed{-88} \).