If `{:A=[(1,0),(k,1)]andB=[(0,0),(k,0)]:}` such that `A^100-I=lambdaB," then "lambda=`

Options

A. 99
B. 100
C. 10
D. 49

Answer

To solve the problem step by step, we will follow the procedure outlined in the video transcript. Here is the detailed solution:

### Step 1: Define the Matrices
We have two matrices:
\[ A = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 0 \\ k & 0 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \]
Calculating the product:
\[ A^2 = \begin{pmatrix} 1 \cdot 1 + 0 \cdot k & 1 \cdot 0 + 0 \cdot 1 \\ k \cdot 1 + 1 \cdot k & k \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2k & 1 \end{pmatrix} \]

### Step 3: Calculate \( A^3 \)
Next, we find \( A^3 \) by multiplying \( A^2 \) by \( A \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 \\ 2k & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \]
Calculating the product:
\[ A^3 = \begin{pmatrix} 1 \cdot 1 + 0 \cdot k & 1 \cdot 0 + 0 \cdot 1 \\ 2k \cdot 1 + 1 \cdot k & 2k \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3k & 1 \end{pmatrix} \]

### Step 4: Calculate \( A^4 \)
Now we calculate \( A^4 \):
\[ A^4 = A^3 \cdot A = \begin{pmatrix} 1 & 0 \\ 3k & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} \]
Calculating the product:
\[ A^4 = \begin{pmatrix} 1 & 0 \\ 3k + k & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 4k & 1 \end{pmatrix} \]

### Step 5: Identify the Pattern
From the calculations, we can see a pattern emerging:
- \( A^2 = \begin{pmatrix} 1 & 0 \\ 2k & 1 \end{pmatrix} \)
- \( A^3 = \begin{pmatrix} 1 & 0 \\ 3k & 1 \end{pmatrix} \)
- \( A^4 = \begin{pmatrix} 1 & 0 \\ 4k & 1 \end{pmatrix} \)

Thus, we can generalize:
\[ A^n = \begin{pmatrix} 1 & 0 \\ nk & 1 \end{pmatrix} \]

### Step 6: Calculate \( A^{100} \)
Using the pattern, we find:
\[ A^{100} = \begin{pmatrix} 1 & 0 \\ 100k & 1 \end{pmatrix} \]

### Step 7: Substitute into the Given Equation
We need to compute \( A^{100} - I \):
\[ A^{100} - I = \begin{pmatrix} 1 & 0 \\ 100k & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 100k & 0 \end{pmatrix} \]

### Step 8: Set the Equation
We have:
\[ A^{100} - I = \lambda B \]
Substituting for \( B \):
\[ \begin{pmatrix} 0 & 0 \\ 100k & 0 \end{pmatrix} = \lambda \begin{pmatrix} 0 & 0 \\ k & 0 \end{pmatrix} \]

### Step 9: Compare Elements
From the equation, we can compare the elements:
- The first row gives no information since both sides are zero.
- From the second row, we have:
\[ 100k = \lambda k \]
Assuming \( k \neq 0 \), we can divide both sides by \( k \):
\[ \lambda = 100 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 100 \]