If `A=[{:(4,1),(3,2):}] and I=[{:(1,0),(0,1):}]` then `A^(2)-6A` is equal to :

Options

A. 3I
B. `-5I`
C. `5I`
D. none of these

Answer

To solve the problem, we need to calculate \( A^2 - 6A \) where \( A = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \) and \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \times A = \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \times \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} \]

Using the formula for matrix multiplication:

\[ A^2 = \begin{pmatrix} (4 \times 4 + 1 \times 3) & (4 \times 1 + 1 \times 2) \\ (3 \times 4 + 2 \times 3) & (3 \times 1 + 2 \times 2) \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 4 \times 4 + 1 \times 3 = 16 + 3 = 19 \)
- First row, second column: \( 4 \times 1 + 1 \times 2 = 4 + 2 = 6 \)
- Second row, first column: \( 3 \times 4 + 2 \times 3 = 12 + 6 = 18 \)
- Second row, second column: \( 3 \times 1 + 2 \times 2 = 3 + 4 = 7 \)

So,

\[ A^2 = \begin{pmatrix} 19 & 6 \\ 18 & 7 \end{pmatrix} \]

### Step 2: Calculate \( 6A \)

Now we calculate \( 6A \):

\[ 6A = 6 \times \begin{pmatrix} 4 & 1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} 6 \times 4 & 6 \times 1 \\ 6 \times 3 & 6 \times 2 \end{pmatrix} = \begin{pmatrix} 24 & 6 \\ 18 & 12 \end{pmatrix} \]

### Step 3: Calculate \( A^2 - 6A \)

Now we subtract \( 6A \) from \( A^2 \):

\[ A^2 - 6A = \begin{pmatrix} 19 & 6 \\ 18 & 7 \end{pmatrix} - \begin{pmatrix} 24 & 6 \\ 18 & 12 \end{pmatrix} \]

Calculating the subtraction element-wise:

- First row, first column: \( 19 - 24 = -5 \)
- First row, second column: \( 6 - 6 = 0 \)
- Second row, first column: \( 18 - 18 = 0 \)
- Second row, second column: \( 7 - 12 = -5 \)

So,

\[ A^2 - 6A = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix} \]

### Step 4: Factor out \(-5\)

We can factor out \(-5\):

\[ A^2 - 6A = -5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = -5I \]

### Final Answer

Thus, the final result is:

\[ A^2 - 6A = -5I \]